SolutionManual InOrganicChemistry by Freeman 2010

Answers to self-tests and exercisesCHAPTER 1 1.7 Take the summation of the rest masses of all the nuclei of the products minus the masses of the nuclei of the reactants. If you get a negative number, energy will be released. But what you have calculated is the mass difference, which in the case of a nuclear reaction is converted to energy. 1.8 0.25 1.9 –13.2 eV 1.10 1524nm, 1.524 X 104 cm-1 Self-tests 1 0 + S1.1 80 35 Br S1.2 d orbitals, 5 orbitals S1.3 4 S1.4 3p S1.5 The added p electron is in a different (p) orbital, so it is less shielded. S1.6 Ni :[Ar]3d 8 4s 2 , Ni 2+ :[Ar]3d 8 S1.7 Period 4, Group 2, s block S1.8 Going down a group the atomic radius increases and the first ionization energy generally decreases. S1.9 81 35 Br n + γ 1.11 1 λ 1 λ 1 Group 16. The first four electrons are removed with gradually increasing values. Removing the fifth electron requires a large increase in energy, indicating breaking into a complete subshell. λ 1 λ ⎛1 1 ⎞ 7 −1 − ⎟ = 1.0974 X10 m 2 ⎝ 1 ∞2 ⎠ = R⎜ ⎛1 1 ⎞ 7 −1 − ⎟ = 1.0288X10 m 2 2 ⎝1 ⎠ 4 = R⎜ ⎛1 1⎞ − ⎟ = 9.7547X106 m −1 ⎝ 12 32 ⎠ = R⎜ ⎛1 1⎞ − ⎟ = 8.2305X106 m −1 ⎝ 12 22 ⎠ = R⎜ S1.10 Adding another electron to C would result in the stable half filled p subshell. 1.12 0 up to n-X S1.11 Cs+ 1.13 n2 1.14 Exercises 1.1 (a) 14 7 N+ 42 He→178 O+11p + γ (b) 12 6 (c) 14 7 C+11p→137 N + γ 1 0 3 1 12 6 N+ n→ H+ C 1.2 246 96 1 Cm+126 C→257 112 Uub+ 0 n 1.3 The higher value of I2 for Cr relative to Mn is a consequence of the special stability of halffilled subshell configurations and the higher Zeff of a 3d electron verses a 4s electron. 4 2 1.5 9 4 1.6 Since helium-4 is the basic building block, most additional fusion processes will produce nuclei with even atomic numbers. 9 4 12 6 l ml Orbital designation 2 1 2p 3 2 3d 5 4 4 0 3 +1, 0, −1 +2, +1, …, −2 0 +3, +2, …, −3 Number of orbitals 3 4s 4f 1 7 1.15 n=5, l = 3, and ml = -3,-2,-1,0,1,2,3 1.16 Li: σ = Z – Zeff ; σ = 3-1.28 = 1.72 Be: σ = Z – Zeff ; σ = 4-1.19 = 2.09 Ne + He → Mg + n 1.4 22 10 N 25 12 1 0 B: σ = Z – Zeff ; σ = 5-2.42 = 2.58 4 2 1 0 C: σ = Z – Zeff ; σ = 6-3.14 = 2.86 Be+ Be→ C+ He +2 n N: σ = Z – Zeff ; σ = 7-3.83 = 3.17 O: σ = Z – Zeff ; σ = 8-4.45 = 3.55 F: σ = Z – Zeff ; σ = 9-5.10 = 3.90 1.17 1 The 1s electrons shield the positive charge form the 2s electrons. ANSWERS TO SELF-TESTS AND EXERCISES 2 1.18 See Figs 1.11 through 1.16 1.19 See Table 1.6 and discussion. 1.20 Table 1.6 shows SR > Ba < Ra. Ra is anomalous because of higher Zeff due to lanthanide contraction. 1.21 Anomalously high value for Cr is associated with the stability of a half filled d shell. 1.22 (a) [He]2s22p2 1.30 (b) [He]2s22p5 (c) [Ar]4s2 (d) [Ar]3d10 (e) [Xe]4f145d106s26p3 (f) [Xe]4f145d106s2 1.23 (a) [Ar]3d14s2 CHAPTER 2 Self-tests S2.1 (b) [Ar]3d2 (c) [Ar]3d5 (d) [Ar]3d4 (e) [Ar]3d S2.2 6 (b) square planar (f) [Ar] 10 (g) [Ar]3d 4s 1 S2.3 Linear S2.4 S22– : 1σg22σu23σg21πu42πg4 ; (h) [Xe]4f 7 1.24 (a) Angular Cl2– : 1σg22σu23σg21πu42πg44σu1. (a) Xe]4f145d46s2 S2.5 1σg22σu23σg21πu42πg4 (b) [Kr]4d6 S2.6 ½[2-2+4+2] = 3 S2.7 Bond order: C≡N, C=N, and C–N; Bond strength: C≡N > C=N > C–N. S2.8 If it contains 4 or fewer electrons. S2.9 –21 kJ mol–1 S2.10 (a) +1/2 6 (c) [Xe]4f (d) [Xe]4f7 (e) [Ar] (f) [Kr]4d 1.25 2 (a) S (b) Sr (b) +5 (c) V (d) Tc Exercises (e) In (f) Sm 2.1 (a) angular 1.26 See Figure 1.4. (b) tetrahedral 1.27 (a) I1 increases across the row except for a dip at S; (b) Ae tends to increase except for Mg (filled subshell), P (half filled subshell), and at AR (filled shell). (c) tetrahedral 1.28 Radii of Period 4 and 5 d-metals are similar because of lanthanide contraction. 1.29 2s2 and 2p0 2.2 (a) trigonal planar (b) trigonal pyramidal (c) square pyramidal 2.3 (a) T-shaped ANSWERS TO SELF-TESTS AND EXERCISES 2.4 (b) square planar (b) one (c) linear (c) none (a) (d) two 2.15 3 (a) 1σg22σu2 (b) 1σg22σu21πu2 Cl Cl (c) 1σg22σu21πu43σg1 I Cl Cl (b) (d) 1σg22σu23σg21πu42πg3 2.16 The configuration for the neutral C2 would be 1σg21σu2 1πu4. The bond order would be ½[22+4] = 2. 2.17 (a) 1σg22σu23σg21πu42πg4 (b) 1 F (c) There is no bond between the two atoms. F S 1800 F 2.5 2.6 2.18 120 0 (a) 2 (b) 1 (c) 2 F 2.19 (a) +0.5 (a) tetrahedral (b) –0.5 (b) octahedral (c) +0.5 2.20 (a) 176 pm (b) 217 pm (c) 221 pm 2.7 2(Si–O) = 932kJ > Si=O = 640kJ; therefore two Si–O are preferred and SiO2 should (and does) have four single Si–O bonds. 2.8 Multiple bonds are much stronger for period 2 elements than heavier elements 2.9 –483 kJ difference is smaller than expected because bond energies are not accurate. 2.10 2.21 (a) 0 (a-c) -1 (b) 205 kJ mol 2.11 Difference in electronegativities are AB 0.5, AD 2.5, BD 2.0, and AC 1.0. The increasing covalent character AD < BD < AC < AB. 2.12 (a) covalent (b) ionic (d) Possibly stable in isolation (only bonding and nonbonding orbitals are filled; not stable in solution because solvents would have higher proton affinity than He. (c) ionic 2.13 (a) sp2 (b) sp3 (c) sp3d or spd3 2 2 2 (d) p d or sp d. 2.14 (a) one 2.22 1 2.23 HOMO exclusively F; LUMO mainly S. 2.24 (a) electron deficient ANSWERS TO SELF-TESTS AND EXERCISES 4 (b) electron precise CHAPTER 3 Self-tests 3.4 3.5 XA2 K3C60 3.6 281 pm 3.7 429 pm 3.8 CuAu. Primitive 12 carat. 3.9 Zintl phase region 3.10 (a) 6:6 and 8:8 S3.1 See Fig. 3.7 and Fig. 3.32. S3.2 See Figure 3.35. S3.3 52% S3.4 rh = ((3/2)1/2 – 1) r = 0.225 r S3.5 409 pm 3.11 6 S3.6 401 pm 3.12 S3.7 FeCr3 2B type and 4A type – in a distorted octahedral arrangement. S3.8 X2A3. 3.13 S3.9 Ti CN = 6 (thought these are as two slightly different distances it is often described as 4 + 2) and O CN = 3. (a) ρ = 0.78, so fluorite (b) FrI? ρ = 0.94, so CsCl (c) BeO? ρ = 0.19, so ZnS (d) InN? ρ = 0.46, so NaCl 3.14 CsCl. S3.10 LaInO3 3.15 S3.11 2421 kJ mol–1 Lattice enthalpies for the di- and the trivalent ions. Also, the bond energy and third electron gain enthalpy for nitrogen will be large. S3.12 Unlikely 3.16 4 four times the NaCl value or 3144 kJmol–1. S3.13 MgSO4 < CaSO4 < SrSO4 < BaSO4. 3.17 (a) 10906 kJmol–1 S3.14 NaClO4 S3.15 Schottky defects. S3.16 Phosphorus and aluminium. S3.17 The dx2-y2 and dz2 have lobes pointing along the cell edges to the nearest neighbor metals. S3.18 (a) n-type (b) p-type Exercises 3.1 a ≠ b ≠ c and α = 90°, β = 90°, γ = 90° 3.2 Points on the cell corners at (0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1), and (1,1,1) and in the cell faces at (½,½,0), (½,1,½), (0,½,½) (½,½,1), (½,1,½), and (1,½,½). 3.3 (c) and (f) are not as they have neighbouring layers of the same position. (b) CsCI (b) 1888 kJ mol–1 (c) 664 kJ mol–1 3.18 (a) MgSO4 (b) NaBF4 3.19 CsI < RbCl < LiF < CaO < NiO < AlN 3.20 Ba2+; solubilities decrease with increasing radius of the cation. 3.21 (a) Schottky defects (b) Frenkel defects 3.22 Solids have a greater number of defects as temperatures approaches their melting points. 3.23 The origin of the blue color involves electron transfer from cationic centres. Vanandium carbide and manganese oxide. 3.24 ANSWERS TO SELF-TESTS AND EXERCISES 3.25 3.26 3.27 Yes. A semiconductor is a substance with an electrical conductivity that decreases with increasing temperature. It has a small, measurable band gap. A semimetal is a solid whose band structure has a zero density of states and no measurable band gap. Ag2S and CuBr: p-type; VO2: n-type. CHAPTER 4 S4.9 5 Identify the acids and bases? (a) FeCl3 + Cl– → [FeCl4]–, acid is FeCl3, base is Cl–. (b) I– + I2 → I3–, acid is I2, base is I–. S4.10 The difference in structure between (H3Si)3N and (H3C)3N? The N atom of (H3Si)3N is trigonal planar, whereas the N atom of (H3C)3N is trigonal pyramidal. S4.11 Draw the structure of BF3·OEt2? Self-tests S4.1 (a) HNO3 + H2O → H3O+ + NO3– HNO3, acid. Nitrate ion, conjugate base. H2O, base. H3O+, conjugate acid. (b) CO32– + H2O → HCO3– + OH– carbonate ion, base; hydrogen carbonate, or bicarbonate, conjugate acid; H2O, acid; hydroxide ion, conjugate base. Exercises 4.1 (c) NH3 + H2S → NH4+ + HS– Ammonia, base; NH4+, conjugate acid; hydrogen sulphide, acid; HS–, conjugate base. S4.2 What is the pH of a 0.10 M HF solution? pH= 2.24 S4.3 Calculate the pH of a 0.20 M tartaric acid solution? Sketch an outline of the s and p blocks of the periodic table, showing the elements that form acidic, basic, and amphoteric oxides? The elements that form basic oxides are in plain type, those forming acidic oxides are in outline type, and those forming amphoteric oxides are in boldface type. pH=1.85 S4.4 Which solvent? dimethylsulfoxide (DMSO) and ammonia. S4.5 Is aKBrF4 an acid or a base in BrF3? A base. S4.6 Arrange in order of increasing acidity? The order of increasing acidity is [Na(H2O)6]+ < [Ni(H2O)6]2+ < < [Mn(H2O)6]2+ 3+ [Sc(H2O)6] . S4.7 4.2 Predict pKa values? (a) H3PO4 pKa ≈ 3. The actual value, given in Table 4.1, is 2.1. of the (b) HSO4–? The conjugate base is SO4–. (b) H2PO4 pKa(2) ≈ 8. The actual value, given in Table 4.1, is 7.4. (c) CH3OH? The conjugate base is CH3O–. (d) H2PO4–? The conjugate base is HPO42–. (c) HPO42– pKa(3) ≈ 13. The actual value, given in Table 4.1, is 12.7. (e) Si(OH)4? SiO(OH)3–. What happens to Ti(IV) in aqueous solution as the pH is raised? Treatment with ammonia causes the precipitation of TiO2. Further treatment with NaOH causes the TiO2 to redissolve. bases (a) [Co(NH3)5(OH2)]3+, conjugate base is [Co(NH3)5(OH)]2+. – S4.8 Identify the conjugate following acids? (f) HS−? 4.3 The conjugate base is The conjugate base is S2–. Identify the conjugate following bases? acids of the H4SiO4.4 Calculate the [H3O+] and pH of a 0. ClO4–.14 - F will behave as a base in water. Na+ or Ag+? Ag+(aq). and ClO4–? The acidity of the four conjugate acids increases in the order HSiO43– < HPO42– < HSO4– < HClO4. 4. SO42–. C5H6N+. the predicted pKa = 3.13 –10 What is the Ka for C5H5NH+? - Predict if F will behave as an acid or a base in water? 4. studied NO3 is of directly measurable base strength in liquid H2SO4. – strong to be 4.17 Which aqua ion is the stronger acid. is more acidic. Kb = 5. BaO. Ka = 5. Pauling’s rules are only approximate. NH3. Cl2O7. not too experimentally. 4. actual value = 2. the predicted pKa = –2. . actual value = –1. (f) CN–? 4. CH3GeH3. and NO3– in water? H2SO4 is a stronger acid. NO3–.ANSWERS TO SELF-TESTS AND EXERCISES 6 ClO4–. is too strong to be studied experimentally in water.6 4. O chloric acid (d) HClO3 or HClO4? HClO4 is a stronger acid. and HClO4? The conjugate acid is HCN. (a) C5H5N (pyridine)? The conjugate acid is pyridinium ion. and ClO4–. O2–.6 × 10 the order is HClO4 > HBrO3 > H2SO4 > HNO2. [Fe(OH2)6]3+.85 4. (b) [Al(OH2)6]3+ or [Ga(OH2)6]3+? O Cl Account for the trends in the pKa values of the conjugate acids of SiO44–. HMnO4 is the stronger acid. (f) H3PO4 or H2SO4? Chlorous acid. H3O+. and SO3? order of increasing basicity is Cl2O7 < SO3 < CO2 < B2O3 < Al2O3 < BaO. ClO42– and NO3– are too weak to be studied experimentally. (b) HSO4–. PO43–. CO2. (c) O2–? The conjugate acid is OH–. HBrO3. is the stronger acid.5 What is the Kb of ethanoic acid? 4.11 Is the pKa for HAsO42– consistent with Pauling’s rules? No.6 × 10–6 4.10 M butanoic acid solution? pH=2. B2O3. Which of the following is the stronger acid? (a) [Fe(OH2)6]3+ or [Fe(OH2)6]2+? The Fe(III) complex. 4. electron withdrawing (d) CH3COOH? The conjugate acid is CH3C(OH)2+. chlorous acid (e) H2CrO4 or HMnO4? Chloric acid. 4.8 What are the structures and the pKa values of chloric (HClO3) and chlorous (HClO2) acid? O O the aluminum-containing species is more acidic. Arrange the following oxides in order of increasing basicity? Al2O3. in H2SO4? HSO4–.9 (b) HPO42–? The conjugate acid is H2PO42–.16 Arrange the following in order of increasing acidity? HSO4−. 4.12 What is the order of increasing acid strength for HNO2.8 Which bases are too strong or too weak to be studied experimentally? (a) CO32– O2–.7 Is the –CN group electron donating or withdrawing? 4. 4. and HSO3F? increasing acidity is NH3 < CH3GeH3 < H4SiO4 < HSO4– < H3O+ < HSO3F. Cl H O (c) Si(OH)4 or Ge(OH)4? H Si(OH)4. cannot be studied in sulfuric acid. (e) [Co(CO)4]–? The conjugate acid is HCo(CO)4.15 CO32– is of directly measurable base strength. H2SO4. BeCl2 or BCl3? Write balanced equations for the formation of P4O124– from PO43– and for the formation of [(H2O) 4Fe(OH)2Fe(OH2)4]4+ from [Fe(OH2)6]3+? 4PO43– + 8H3O+ → P4O124– + 12H2O 4.31 Write a balanced equation dissolution of SiO2 by HF? for the . so it will bond more strongly with the softer Lewis acid BH3 H2SO4 + HF ⇔ H3SO4+ + F- 4. As. More balanced equations? (a) H3PO4 and Na2HPO4? 4. Identifying elements that form Lewis acids? All of the p-block elements except nitrogen. Choose between the two basic sites in Me2NPF2? The phosphorus atom in Me2NPF2 is the softer of the two basic sites. (b) Me[B12]– + Hg2+ → [B12] + MeHg+? The Lewis acid Hg2+ displaces the Lewis acid [B12] from the Lewis base CH3–. Mo.19 The change in charge upon aqua ion polymerization? 4.18 Which of the following elements form oxide polyanions or polycations? Al. [(H2O)4Fe(OH)2Fe(OH2)4+ + 2H3O+ Ca2+ + Boron trichloride. 4.27 – 4. and the lighter noble gases form Lewis acids in one of their oxidation states.+ H2F+ Why is H2Se a stronger acid than H2S? As you go down a family in the periodic chart. B. (b) More basic toward BMe3: NMe3 or NEt3? NMe3. The ambiguity for DMSO is that both the oxygen atom and sulfur atom are potential basic sites.22 B(n-Bu)3. 4.23 Which of the following reactions have Keq > 1? (a) R3P–BBr3 + R3N–BF3→ R3P–BF3 + R3N–BBr3? <1 (c) CH3HgI + HCl → CH3HgCl + HI? H3PO4 + HPO42. (c) KCl + SnCl2 → K+ + [SnCl3]–? The Lewis acid SnCl2 displaces the Lewis acid K+ from the Lewis base Cl–. polycations: Al. fluorine. either base could be stronger.24 <1 (d) [AgCl2] (aq) + 2 CN (aq) → [Ag (CN)2]– (aq) + 2Cl–(aq)? >1 (b) CO2 and CaCO3? 4.26 Polycation formation reduces the average positive charge per central M atom by +1 per M.25 – 2HCO 3- Give the equations for HF in H2SO4 and HF in liquid NH3? 4. BCl3.29 Why is trimethylamine out of line? Trimethyl amine is sterically large enough to fall out of line with the given enthalpies of reaction.30 Discuss relative basicities? (a) Acetone and DMSO? DMSO is the stronger base regardless of how hard or how soft the Lewis acid is. (e) EtOH readily dissolves in pyridine? A Lewis acid–base complex formation reaction between EtOH (the acid) and py (the base) produces the adduct EtOH–py.21 7 (b) Me2S and DMSO? Depending on the EA and CA values for the Lewis acid. B(n-Bu)3 or B(t-Bu)3? 2[Fe(OH2)6]3+ → CO 2 + CaCO 3 + H 2O Select the compound with the named characteristic? (a) Strongest Lewis acid: BF3. or BBr3? BBr3.ÅÆ 2H2PO4- NH3 + HF 4-Me-py. Ti? (d) AsF3(g) + SbF5(g) → [AsF2][SbF6]? The very strong Lewis acid SbF5 displaces the Lewis acid [AsF2]+ from the Lewis base F–. 4. The hard nitrogen atom will bond more strongly to the hard Lewis acid BF3. 4. polyoxoanions (oxide polyanions): Mo polyoxoanions: As. 4. Cu. Cu. (b) SO2 + Ph3P–HOCMe3 → Ph3P–SO2 + HOCMe3? > 1. Si.20 2-Me-py or 4-Me-py? 4. Identifying acids and bases: (a) SO3 + H2O → HSO4– + H+? The acids in this reaction are the Lewis acids SO3 and H+ and the base is the Lewis base OH–. the acidy of the homologous hydrogen compounds increases. oxygen.28 NH2. B. and Si.ANSWERS TO SELF-TESTS AND EXERCISES 4. and Ti. Cl– couple? Eox= – 1.10 Frost diagram for thallium in aqueous acid? The balanced equation CH 3CH 2OH 2+ + F- (b) NH3? The equation is NH NH oxidation 3+ (d) Promote the reaction 2FeCl3 + ZnCl2 →Zn2+ + 2[FeCl4]−? A suitable solvent is acetonitrile.+ Cl- (b) Favor basicity of R3As over R3N? Alcohols such as methanol or ethanol would be suitable. S5.33 CHAPTER 5 Self-tests S5.6 Can Fe2+ disproportionate under standard conditions? No.37 The dissolution of silicates by HF? both 4.+ I- ΔfH = -20.+ H2F+ 4 + + F- .0 bar? E = 1. S5. S5. 4.0kJ/mol Al2 O 3 + 3H 2 S Al2 S 3 + 3H 2 O 4.36 (a) CH3CH2OH? is: CH 3CH 2OH + HF 3 + HF 2MnO4− (aq) + 5Zn(s) + 16H+(aq) → 5Zn2+(aq) + 2Mn2+(aq) + 8H2O(l) S5. 4.25 V.38 Are the f-block elements hard? yes.32 (c) C6H5COOH? 4. Write Brønsted acid–base reactions in liquid HF? 4.8 Potential of AgCl/Ag. S5. 4. would provide Lewis acidic sites that could abstract Cl–: Why does Hg(II) occur only as HgS? Mercury(II) is a soft Lewis acid. MeCN.3 Can Cr2O72– be used to oxidize Fe2+.1 acid-I.34 C6H5COO.oxidation is not a problem. S5. the most common of which is S2–.7 bpy binding to Fe(III) or Fe(II)? Fe(II) preferentially. + Propose a mechanism for the acylation of benzene? An alumina surface. S5. 4. such as the partially dehydroxylated one shown below. Cl.2 Does Cu metal dissolve in dilute HCl? No. Another suitable solvent is H2O.ANSWERS TO SELF-TESTS AND EXERCISES 8 SiO2 + 6HF 2H2O + H2SiF6 or SiO2 + 4HF C6H5COOH + HF 2H2O + SiF4 Both a Brønsted acid–base reaction and a Lewis acid–base reaction.35 Half-reactions and balanced reaction for oxidation of zinc metal by permanganate ions? 2[MnO4− (aq) + 8H+ (aq) + 5e– → Mn2+(aq) reduction + 4H2O(l)] 5 [ Zn(s) → Zn2+(aq) + 2e– ] (c) Favor acidity of Ag over Al ? An example of a suitable solvent is diethyl ether. and would Cl– oxidation be a problem? Yes.5 The fate of SO2 emitted into clouds? The aqueous solution of SO42– and H+ ions precipitates as acid rain. you can shift the following equilibrium to the right: acid-Cl.4 Fuel cell emf with oxygen and hydrogen gases at 5.39 Calculate the enthalpy change for I2 with phenol? Write a balanced equation to explain the foul odor of damp Al2S3? The foul odor suggests H2S formation. φ Describe solvent properties? (a) Favor displacement of Cl– by I– from an acid center? If you choose a solvent that decreases the activity of chloride relative to iodide.38 V S5. (b) Pu(V) does not disproportionate into Pu(VI) and Pu(IV). and so is found in nature only combined with soft Lewis bases.9 Latimer diagram for Pu? (a) Pu(IV) disproportionates to Pu(III) and Pu(V) in aqueous solution. S5. S5.1 (iv) disproportionation will not occur. disproportionate in aqueous acid to Br– and HBrO. Ru2+ will disproportionate in aqueous acid to Ru3+ and metallic ruthenium.13 S5. HClO2 will reduce O2 and in doing so will be oxidized to ClO3–.4 Balanced equations for redox reactions? (a) Fe2+? (i) Fe2+ will not oxidize water. The minimum temperature for reduction of MgO by carbon? 1800ºC or above. Nitrate is a stronger oxidizing agent in acidic solution than in basic solution. 5. (c) Reducing Ag+(aq) to Ag(s)? The reduced form of any couple with a reduction potential less than 0.ANSWERS TO SELF-TESTS AND EXERCISES 5. HClO2 will disproportionate in aqueous acid to ClO3– and HClO.99 V +2H2O(l) – Compare the strength of NO3 as an oxidizing agent in acidic and basic solution? A competing reaction is: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) (Eº = 0. (d) HOCl? No reaction. Ru2+ will reduce O2 and in doing so will be oxidized to Ru3+. H2O2. LiCoO2(s) + C(s) → LiC(s) + CoO2(s) +1 +3 -2 0 +1-1 +4 -2 Ca(s) + H2(g) 0 0 5. 2Mn3+(aq) + 2H2O → MnO2 + Mn2+ + 4H+(aq) +3 +1 -2 +4 -2 +2 +1 (c) HClO2? HClO2 will oxidize water.535 V.12 2Zn(s) + O2(g) + 4H+(aq) → 2Zn2+(aq) Eº = 1.6 Balance redox reaction in acid solution: MnO4– + H2SO3 → Mn2+ + HSO4–? pH dependence? . metallic zinc. (d) Reducing I2 to I–? The reduced form of any couple with a reduction potential less than 0. Oxidation numbers? 2 NO(g) + O2(g) → 2 NO2(g) +1 -2 0 +4 -2 (b) Ru2+? Ru2+ will not oxidize or reduce water. or α–PbO2 to oxidize Cl– to Cl2. (d) Br2? Br2 will not oxidize or reduce water. will not reduce water. if a reaction occurs. or NH3OH+. Exercises (iii) Fe2+ will reduce O2 and in doing so will be oxidized to Fe3+.5 Standard potentials vary with temperature in opposite directions? The amino and cyano complexes must have different equilibrium shifts with respect to changes in temperature that results in the opposite directions of change for the cell potential. S5. for the following species in aerated aqueous acid? (a) Cr2+? 4Cr2+(aq) + O2(g) + 4H+(aq) → 4Cr3+(aq) Eº = 1.11 The oxidation number of manganese? (e) Zn(s)? Mn2+(aq) S5.763 V).799 V.46 V + 2H2O(l) (c) Cl–? no reaction.14 The possibility of finding Fe(OH)3 in a waterlogged soil? Fe(OH)3 is not stable.3 9 Write balanced equations.65 V + 2H2O(l) (b) Fe2+? 4Fe2+(aq) + O2(g) + 4H+(aq) → 4Fe3+(aq) Eº = 0. (b) Reducing Cr3+(aq) to Cr2+(aq)? metallic manganese. 5. 5. Br2 will not Br2 will not reduce O2. (ii) Fe2+ will not reduce water.2 → CaH2(s) +2 -1 Suggest chemical reagents for redox transformations? (a) Oxidation of HCl to Cl2? S2O82–. 5. Thermodynamic tendency of HO2 to undergo disproportionation? E = +1.23 Which of the boundaries depend on the choice of [Fe2+]? Any boundary between a soluble species and an insoluble species will change as the concentration of the soluble species changes. HO2 disproportionation. and predict the predominant species? (a) Fe? 0.24 Under what conditions will Al reduce MgO? Above about 1400ºC. will undergo (is positive). no difference.10 Electrode potential for Ni2+/Ni couple at pH = 14? E =– 0.387 V. Hg22+ is not likely to undergo disproportionation.11 Will acid or base most favour the following half-reactions? (a) Mn2+ → MnO4–? Base (b) ClO4– → ClO3–? Acid (c) H2O2 → O2? Base (d) I2 → 2I–? Acid or base. Dissolved carbon dioxide corrosive towards iron? Carbon dioxide and water generate carbonic acid which encourages the corrosion process by lowering solution pH. 5.275 V.22 How will edta4– complexation affect M2+ → M0 reductions? The reduction of a M(edta)2– complex will be more difficult than the reduction of the analogous M2+ aqua ion.12 Determine the standard potential for the reduction of ClO4– to Cl2? 1.16 Equilibrium constant for the reaction Pd2+(aq) + 4 Cl–(aq) ≡ [PdCl4]2–(aq) in 1 M HCl(aq)? K = 4. 5. and E = Eº – [(0.8 pH) 5. At pH 14. SO42– would again predominate. and between the two insoluble species.19 5.14 Find the approximate potential of an aerated lake at pH = 6.1 V.7 5. 5. Reduction potential for MnO4– to MnO2(s) at pH = 9.059V)/4][log(1/(p(O2)[H+]4)] (b) The reduction of Fe2O3(s)? 5. (c) Should HClO3 disproportionate in aqueous acid solution? Kinetic.8 (b) What happens when Cl2 is dissolved in aqueous acid solution? Cl2 will not disproportionate.7 × 1038 5. . 0. + 6 Q = 1/[H ] and E = Eº – (RT/nF)(13.20 5. 5. HSO4– is the predominant sulfur species at pH 6.18 Tendency of mercury species to act as an oxidizing agent. so a solution of Cl– and ClO– is formed when Cl2 is dissolved in aqueous base.5 – 0. 5.98 V 5.21 V 5.37 × 1010 5. Cl2 is thermodynamically capable of oxidizing water.17 The oxidation of ClO– is slow. a reducing agent. What is the maximum E for an anaerobic environment rich in Fe2+ and H2S? –0. The boundaries between the two soluble species. 5.55 V Q = 1/(p(O2)[H+]4) (c) S? At pH 0.6 V Write the Nernst equation for (a) The reduction of O2? (b) Mn? E = 0.00? E = 0.ANSWERS TO SELF-TESTS AND EXERCISES 10 2MnO4− (aq) + 5H2SO3(aq) + H+(aq) → 2Mn2+(aq) +5HSO3− (aq) + 3H2O(l) The potential decreases as the pH increases.15 Frost diagram and standard potential for the HSO4−/S8(s) couple? 0. will not depend on the choice of [Fe2+].387 V Using Frost diagrams? (a) What happens when Cl2 is dissolved in aqueous basic Cl2 is thermodynamically solution? susceptible to disproportionation to Cl– and ClO4– when it is dissolved in aqueous base.392 V 5.9 Write equations for the following reactions: (a) N2O is bubbled into aqueous NaOH solution? 5N2O(aq) + 2OH–(aq) → 2NO3–(aq) + 4N2 (g) + H2O(l) (b) Zinc metal is added to aqueous acidic sodium triiodide? – 2+ – Zn(s) + I3 (aq) → Zn (aq) + 3I (aq) (c) I2 is added to excess aqueous acidic HClO3? 3I2(s) + 5ClO3– (aq) + 3H2O(l) → 6IO3– (aq) + 5Cl– (aq) + 6H+(aq) 5. None of these species are likely to be good reducing agents.13 Calculate the equilibrium constant for Au+(aq) + 2CN–(aq) → [Au(CN)2]–(aq)? K = 5.21 5. or to undergo disproportionation? Hg2+ and Hg22+ are both oxidizing agents. 3 Assigning point groups: (a) NH2Cl? Cs (b) CO32–? D3h (c) SiF4? Td (d) HCN? C∞v. S6.15 +T1u. dxy is B2g.11 Orbital symmetry for a square-planar array of H atoms? B2g. stretching vibrations leads to: Γstr = 3A1 (IR and Raman. SALCs for sigma bonding in O? A1g + Eg Exercises Symmetry elements? (a) a C3 axis and a σv plane in the NH3 molecule? N H N H H H H H σv C3 (b) a C4 axis and a σh plane in the squareplanar [PtCl4]2– ion? Cl Pt Cl Cl Cl Cl Cl C4 6.1 Is the skew form of H2O2 chiral? Yes. 6.8 Confirm that the symmetric mode is Ag? D2h character table. and 5px and 5py have Eu symmetry.13 + 2B1 (Raman) + E (IR.3 S6.4 S6. 5s and 4dz2 have A1g symmetry. (c) BF3? neither. which is the Ag symmetry type. S6. S6.1 S6. Fig. SF5Cl has C4v symmetry. Show that the four CO displacements in the square-planar (D4h) [Pt(CO)4]2+ cation transform as A1g + B1g + Eu.2 Pt Cl σh S4 or i? (a) CO2? i (b) C2H2? i.2 Γstr = A1g (Raman. Eu is IR active.7 Can the bending mode of N2O be Raman active? Yes. (Raman) + T1u (IR).10 Orbital symmetry for a tetrahedral array of H atoms in methane? A1 S6. S6. How many bands would you expect in the IR and Raman spectra for the [Pt(CO)4]2+ cation? The reducible representation: D4h E 2C4 C2 2C2’ 2C2″ i 2S4 σh 2 σv 2 σd Γ3N 4 0 0 2 0 0 0 4 2 0 Reduces to A1g + B1g + Eu A1g + B1g are Raman active. 2– (b) SO4 point group? Td. S6. dxz and dyz are Eg. 6. Symmetry species of all five d orbitals of the central Xe atom in XeF4 (D4h.ANSWERS TO SELF-TESTS AND EXERCISES 11 SF6 has Oh symmetry. the dx2-y2 has B1g symmetry. dz2 is A1g.3)? dx2-y2 is B1g.9 Analysis of the Predict how the IR and Raman spectra of SF5Cl differ from that of SF6? S6.14 Symmetries of all the vibration modes of [PdCl4]2-? A1g + B1g + B2g + A2u + B2u + 2Eu A conformation of the ferrocene molecule that lies 4 kJ mol–1 above the lowest energy configuration is a pentagonal antiprism. What is the maximum possible degeneracy for an Oh molecule? 3. (f) BrF4–? D4h. S6. (d) SO42–? three different S4. How many of these axes are there in the ion? Three S4 axes. (e) SiFClBrI? C1. polarized) (a) BF3 point group? D3h.5 S6. Is it polar? No.12 Which Pt atomic orbitals can combine with which of these SALCs? The atomic orbitals much have matching symmetries to generate SALCs. polarized) + Eg Sketch the S4 axis of an NH4+ ion. S6. Analysis of the stretching vibrations leads to: CHAPTER 6 Self-tests S6.6 S6. Cl . 6. Raman). (c) A dxy orbital? Center of symmetry. (d) A dz2 orbital? In addition to the symmetry elements possessed by a p orbital: (i) a center of symmetry. vibrations are: A1’ (Raman.15 Use the projection operator method to construct the SALCs of A1 + T2 symmetry that derive from the four H1s orbitals in methane. The symmetry elements of orbitals? (a) An s orbital? Infinite number of Cn axes.13 (b) A p orbital? An infinite number of mirror planes that pass through both lobes and include the long axis of the orbital. (c) Which atomic orbitals on C can form MOs with H1s SALCs? Using symmetry Td. 6. three mutually perpendicular mirror planes of symmetry.10 IR and Raman to distinguish between: (a) planar and pyramidal forms of PF3.11 s = (1/2)(ϕ1 + ϕ2 + ϕ3 + ϕ3) (= A1) AsCl5 Raman spectrum consistent with a (b) Perpendicular to the molecular plane? 1 6. polarized) + B1 (Raman) + 2B2 (IR and Raman ) + 3E (IR and Raman). (b) planar and 90o-twisted forms of B2F4 (D2h and D2d respectively)? (a) Planar PF3. vibrations are: 2A1 (IR and Raman. i.ANSWERS TO SELF-TESTS AND EXERCISES 12 6. The MOs would be constructed from SALCs with H1s and 2s and 2p atomic orbitals on C. (b) BF3? The E′ modes are active in both IR and Raman. (iii) an infinite number of C2 axes that pass through the center of the orbital and are perpendicular to the C∞ axis. Γ3N reduces to: A1 + T2. SALCs for σ-bonds (a) BF3? (1/√3)(ϕ1 + ϕ2 + ϕ3) (= A1’) (1/√6)(2ϕ1 – ϕ2 – ϕ3) and (1/√2)(ϕ2 – ϕ3) (= E’) (b) PF5? (axial F atoms are ϕ4 + ϕ5) (1/√2)(ϕ4 + ϕ5) (= A1’) (1/√2)(ϕ4 − ϕ5) (= A2”) (1/√3)(ϕ1 + ϕ2 + ϕ3) (= A1’) (1/√6)(2ϕ1 – ϕ2 – ϕ3) and (1/√2)(ϕ2 – ϕ3) (= E’) . Pyramidal PF3.6 For the 90o-twisted form of B2F4 (D2d) The vibrations are: 3A1 (Raman. the long axis is a Cn axis. (ii) a mirror plane that is perpendicular to the C∞ axis.and C6H3Cl3.8 6. plus an infinite number of mirror planes of symmetry. plus center of inversion. D3h. In addition. SO32– ion? (a) Point group? C3v (c) Which s and p orbitals have the maximum degeneracy? 3px and 3py orbitals are doubly degenerate.9 px = (1/2)(ϕ1 – ϕ2 + ϕ3 – ϕ3) (= T2) trigonal bipyamidal geometry? No.14 (a) Take the 4 hydrogen 1s orbitals of CH4 and determine how they transform under Td. where n can be any number from 1 to ∞. PF5? (a) Point group? D3h. C3v. and (iv) an S∞ axis. 6.7 [AuCl4]− ion? Γ of all 3N displacements Vibrations of a C6v molecule that are neither IR nor Raman active? Any A2. (b) Degenerate MOs? 2 6. 6. polarized) + 2E’ (IR and Raman) + A2” (IR). (b) Confirm that it is possible to reduce this representation to A1 + T2. py = (1/2)(ϕ1 – ϕ2 – ϕ3 + ϕ3) (= T2) Vibrational modes of SO3? (a) In the plane of the nuclei? 5 pz = (1/2)(ϕ1 + ϕ2 – ϕ3 – ϕ3) (= T2) Vibrations that are IR and Raman active? (a) SF6? None.4 6. (b) Degenerate MOs? 2. three mutually perpendicular C2 axes. polarized) + 2B2g (Raman) + B3g (Raman) + Au(inactive) + 2B1u (IR) + B2u (IR) + 2B3u (IR). B1. or B2 vibrations of a C6v molecule will not be observed in either the IR spectrum or the Raman spectrum. 6. 6.5 How many planes of symmetry does a benzene molecule possess? What chlorosubstituted benzene has exactly four planes of symmetry? 7.. two planes that are rotated by 45º about the z axis from the xz plane and the yz plane. (c) Which p orbitals have the maximum degeneracy? 3px and 3py atomic orbitals are doubly degenerate 6.12 and irreducible representations? A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu 6. polarized) + 2E’ (IR and Raman) (b) For the planar form of B2F4 (D2h): The vibrations are: 3Ag (Raman. Name and draw the structures of the Nickel complexes? (a) [Ni(CO)4]? tetracarbonyl or tetracarbonyl nickel(0).1 Give formulas corresponding to the following names? (a) Cisdiaquadichloroplatinum(II)? cis[PtCl2(OH2)2]. Kf5 = 810.2 CO CO 2- Cl What type of isomers are possible for [Cr(NO2)2•6H2O]? The hydrate isomers and linkage isomers of the NO2 group. (d) [Mn(NH3)6]2+? Hexaamminemanganesium (II) 2+ S7. they exhibit the same chemical shift. S7. OC (b) Diamminetetra(isothiocyanato) chromate(III)? [Cr(NCS)4(NH3)2] –.3 Name the following complexes? (c) cis-[RhH(CO)(PR3)2]? Not chiral. – (b) trans-[CrCl2(ox)2]3 ? Not chiral. [Cr(ONO)(H2O)5]NO2 •H2O. cis[Cr(NCS)4(NH3)2] – or trans[Cr(NCS)4(NH3)2] –. Kf4 = 2700. Kf3 = 9000.ANSWERS TO SELF-TESTS AND EXERCISES 13 Exercises CHAPTER 7 7.4 Sketches of the mer and fac isomers of [Co(gly)3]? H3N NH3 H3N 7. trans[PtCl2(OH2)2]. . (b) trans-[Cr(NCS)4(NH3)2] ? transdi(ammine)tetrakis(isothiocyanato)chromate (III) - (c) [Co(C2O4)(en)2]+? bis(ethylenediamine)oxalatocobalt(III). (c) Tris(ethylenediamine)rhodium [Rh(en)3]3+. Also.3 Identifying isomers? Note that the two phosphine ligands in the trans isomer are related. CO Ni trans-diaquadichloroplatinum(II). 2- NC (III)? Ni NC (d) Bromopentacarbonylmanganese (I)? [MnBr (CO)5]. and finally Kf6 = 243. S7. (b) [Ni(CN)4]2–? Tetracyanonickelate (II). therefore. Kf2 will be 30% less or 30000. (a) cis-[CrCl2(NH3)4]+? cistetra(ammine)di(chloro)chromium(III) Calculate all of the stepwise formation constants? Kf1 = 1 X 105.2 Mn NH3 NH3 NH3 Write the formulas for the following complexes? (a) [CoCl(NH3)5]Cl2 (b) [Fe(OH2)6](NO3)3 (c) cis-[FeCl2(en)2] S7. Co Cl Cl Cl S7.6 (d) [Cr(NH3)5μ–OH–Cr(NH3)5]Cl5 7.1 Self-tests S7. . CN CN (c) [CoCl4]2–? Tetrachlorocobaltate (II) (e) Chlorotris(triphenylphosphine)rhodium (I)? [RhCl(PPh3)3].can exist as.5 Which of the following are chiral? (a) cis[CrCl2 (ox)2]3–? Chiral . (c) [IrHCO(PR3)2] has two isomers.ANSWERS TO SELF-TESTS AND EXERCISES 14 7. no. two isomers. a bidentate ligand can bond through two atoms.6 Six-coordinate complexes? (a) Sketch the two observed structures? [Mg(edta)(OH2)]2– 7. 7. for a squareplanar complex. sketch the two observed structures? A A B B E M M E B E B A Trigonal Bipyramidal Square based pyramid A = axial ligands A = axial B = basal E = equatorial ligands 7. (b) ox. and quadridentate? A monodentate ligand can bond to a metal atom only at a single atom. and (d) edta? .5 For five-coordinate complexes. Which ligand could act like a chelating ligand? (a) Triphenylphosphite. (d) Pyrazine. (d) [Pd(gly)2] has two isomers. (b) Which one of these is rare? Trigonal prism. no isomers. bidentate.9 7. 7.4 Four-coordinate complexes? (a) Sketch the two observed structures? L M M L L L L L L L square planar tetrahedral (b) Isomers expected for MA2B2? tetrahedral complex.14 How many isomers are possible for the following complexes? (a) [FeCl(OH2)5]2+? None. (c) Bipyridine (bipy). Draw structures of complexes that contain the ligands (a) en.8 7. What type of isomers do you get with ambidentate ligands? linkage isomers. (c) phen. a quadridentate ligand can bond through four atoms. 7.7 7. yes. (b) [IrCl3(PEt3)2]? 2 (c) [Ru(biby)3]2+? 2 (d) [CoCl2(en)(NH3)2]+? 4 (e) [W(CO)4(py)2] 2 . no/ (b) Bis(dimethyl)phosphino ethane (dmpe) yes.10 Explain the difference between monodentate. cis and trans.12 Which complexes have isomers? [CoBrClI(OH2)] 7.13 Which complexes have isomers? (a) [Pt(ox)(NH3)2] no isomers (b) [PdBrCl(PEt3)2] has two isomers. 7.11 What types of isomers are [RuBr(NH3)5]Cl and [RuCl(NH3)5]Br? Ionization isomers. 5. shown in the exercise? The Λ isomer. 8. 4 total vibrational modes. Λ or Δ.7 (c) cis-[RhCl2(NH3)4]+? not chiral.3 What is the minimum size of a cubic crystal that can be studied? 0. B E Which of the following complexes are chiral? (a) [Cr(ox)3]3–? Chiral S8. N N N N Ru N N N N Ru N N N N Λ isomer Δ isomer 7.19 Suggest a reason why Kf5 is so different? Because of a change in coordination. the signal is split into 2 lines.19 and suggest why they are different? The chelate effect.8 Raman bands assignments? N(SiH3)3 is planar.ANSWERS TO SELF-TESTS AND EXERCISES 15 7.2 Why are there no diffraction maxima in borosilicate glass? Glass has no long-range periodicity or order.5 Order of stretching frequencies? The smaller effective mass of the oscillator for CN− causes the molecule to have the higher stretching frequency. 8. 7. The bond order for NO is 2. S8. which has I = ½. 8. 8.5 μm by 0. CHAPTER 8 Self-tests S8. hence CO has a higher stretching frequency than NO.17 Which isomer.1 How would you determine crystalline components in mineral sample? Powder Xray diffraction. Λ or Δ. (b) Proton resonance of the hydrido ligand consist of eight equal intensity lines? yes. Why does the mass spectrum of ClBr3 have five peaks separated by 2 u? Halogen isomers.5 A 14% of naturally occurring tungsten is 183W. and N is heavier than C. 8. thus the long progression. of the complex [Ru(en)3]+2 ? 8. TiO2 in sunscreens? Titania articles absorb this ultraviolet radiation Molecular shape and vibrational modes for XeF2? Trigonal bipyramidal. Exercises (e) fac-[Co(NO2)3(dien)]? Not chiral. (a) 77Se-NMR spectrum consists of a triplet of triplets? The triplet of triplets.20 km/s? 1.2 A D A B M identical reflections to those of rutile TiO2 but shifted to slightly higher diffraction angles.4 E A D S8.1 Main features of the CrO2 powder XRD pattern? XRD pattern for CrO2 will show .15 Draw all possible isomers for [MA2BCDE]? Including optical isomers. is the complex Mn(acac)3.80 × 10–12 m or 180 pm. Thus.5 μm by 0.20 Compare these values with those of ammonia given in exercise 7.4 Wavelength of neutron at 2.5 μm. EPR signal of new material arises from W sites? Chiral (dien is 8.18 Draw both isomers. B M S8. N(CH3) 3 is pyramidal.16 A C A E S8. 8. 7. The ionised molecule has greater planarity. (f) mer-[Co(NO2)3(dien)]? not planar). 7. 15 isomers are possible! . A B C M A A B D E E D E B M C B B D C E C A E B M D D C A A A A D E M A C C D S8.3 M A M A M A C 7. (d) [Ru(bipy)3]2+? chiral Isomer shift for iron in Sr2FeO4? The Smaller and less positive.7 UV photoelectron spectrum of NH3? The band at 11 eV is due to the lone pair and the pyramidal angle.6 Wavenumber for O–O in O2+? In the region of 1800 cm 1.6 (b) cis-[PtCl2(en)]? Chiral (The en is not planar). 94. Exercises 9. S9.22 Isomer shift for iron in BaFe(VI)O4? A positive shift for Fe(VI) well below +0.56.82%) and 109Ag (48. 200. 8.15 8.2 Form saline hydrides. molecular tumbling removes the effect of the g-value anisotropy.ANSWERS TO SELF-TESTS AND EXERCISES 16 8. and in this case the 19F resonance is a doublet. S9. 1. 9. Above 0. In frozen solution.4 Comment on ΔfHө values? It is evident from the values that as we move down the group. The 77Se-NMR spectrum is a triplet of triplets. 107Ag (51. +5. +2. (c) P.18%). and 1.nH2O. Charge on Fe atoms in Fe4[Fe(CN)6]3? EPR and Mössbauer.18 8.9 Single 13C peak in NMR? Chemically distinct carbonyls are exchanging position sufficiently quickly.21 V. No quadrupole splitting in Mössbauer spectrum of SbF5? The geometry must be close to cubic in the solid state.12 g-values? 1. 9.5 Further data useful when drawing comparisons with the value for V2O5? We would have to know the products formed upon decomposition. 8.16 8. gvalue anisotropy can be observed. CHAPTER 9 Self-tests . (d) Cl. and all the carbides react with water to liberate a hydrocarbon? the alkaline earth metals or Group 2 elements. steric crowding of the fluorines is minimized. 8. +7. and 174.2 As an integer. No peak in the mass spectrum of Ag at 108 u? Two isotopes. I = 1/2. S9. Compounds that contain silver will have two mass peaks.5% intensity from the 19F coupled to the 129Xe. oxides and peroxides. S9. The same structure is SmO4. Rb and Sr can be found in aluminosilicate minerals. S9. determine n. it is more likely that polyoxygen anions will form pi bonds that limit extended bonding owing to restrictions on pi orbital overlap through multiple bridging centres.74. 9.17 8. 230.? n=7. 8. form halides in oxidation states +5 and +3 and toxic gaseous hydrides? Elements in Group 15. NMR spectral features for XeF5−? All 5 of the F atoms are chemically equivalent. n = 7.11 8. 9.13 Slower process. Zeolite of composition CaAl2Si6O16.20 8. Peaks in mass spectrum of Mo(C6H6)(CO)3? 258. 186.2 Sulfur forms catenated polysulfides whereas polyoxygen anions are unknown? Owing to a strong tendency to form strong double bonds.2 mm s-1.21 8.1 Found in aluminosilicate minerals or sulfides? Cd and Pb will be found as sulfides.4 Born–Haber cycle for the formation of the hypothetical compound NaCl2? Which thermochemical step is responsible for the fact that NaCl2 does not exist? The second ionization energy of sodium is 4562 kJ mol-1 and is responsible for the fact that the compound does not exist.19 8. Cyclic voltammogram of Fe(III) complex? The complex undergoes a reversible oneelectron reduction with a reduction potential of 0. NMR or EPR? NMR. Ratio of cobalt to acetylacetonate in the product? The ratio is 3:1. Cr and Pd can be found in both oxides and sulfides. The final result is a composite: two lines of 12. +5.3 Elements vary from metals through metalloids to non-metals.14 Differences in EPR spectrum for d-metal with one electron in solution versus frozen? In aqueous solution at room temperature.5 Inert pair effect beyond Group 15? The relative stability of an oxidation state in which the oxidation number is 2 less than the group number is an example of the inert pair effect.3 Shape of XeO4 and identify the Z + 8 compound with the same structure? A tetrahedral geometry. and one remaining line of 75%. (b) As. Approximately 25% is present as 129Xe. 8.720 V the complex is oxidized.10 Form of 19F-NMR and 77Se-NMR spectra of 77SeF4? 19F NMR spectrum reveals two 1:3:3:1 quartets.1 Maximum stable oxidation state? (a) Ba. (c) LiOH(s) + H2(g) → NR.10 Phases of hydrides of the elements? BaH2 and PdH0. 10. Re = +7. and metallic character? Metallic character.4 Preparation of hydrogen gas? (i) CH4(g) + H2O → CO(g) + 3H2(g) (1000°C) (ii) C(s) + H2O → CO(g) + H2(g) (1000°C) (iii) CO(g) + H2O → CO2(g) + H2 (g) CHAPTER 10 Self-tests 10. or GeH4 would best H+ or H donor? CH4. ionization energy. 10. but chemically it fits well in both group 1 and group 17. 10.3 A procedure for making Et3MeSn? 2Et3SnH + 2Na → 2Na+Et3Sn– + H2 Na+Et3Sn– + CH3Br → Et3MeSn + NaBr (a) BaH2? barium hydride.1 (c) NH3? ammonia.8 Identify the Z + 8 element for P. Exercises (b) SiH4? silane. ionic radii decrease across a period and down a group. (d) AsH3? Arsine.ANSWERS TO SELF-TESTS AND EXERCISES 17 9. (d) H2SO4? H = +1.8 Name and classify the following? (b) NH3(g) + BF3(g) → H3N–BF3(g). 10. The halogens are diatomic gases just like hydrogen. S10. (e) PdH0. K = +1.6 What are the physical properties of water without hydrogen bonding? It most likely would be a gas at room temperature.2 Low reactivity of hydrogen? Hydrogen exists as a diatomic molecule (H2). H bonded to the phosphorus atom? If they are assigned an oxidation number of +1.7 Which molecule has the stronger hydrogen bonds? S–H···O has a weaker hydrogen bond than O–H···S. (e) H2PO(OH)? H bonded to an oxygen atom = +1.3 (a) H2S? H = +1.9 Calculate ΔfHө for SeF6? ΔfHө = −1397 kJ mol-1. S = +6.9 . and O = –2.5 Properties of hydrides of the elements? (a) Position in the periodic table? See Figure 10. (c) Variable composition? PdH0. (f) HI? hydrogen iodide.9 (a) Hydridic character? Barium hydride (b) Brønsted acidity? Hydrogen iodide. P2H4.11 Structures of H2Se.6 Ionic radii. would be the best hydride donor. Where does Hydrogen fit in the periodic chart? (a) Hydrogen in group 1? Hydrogen has one valence electron like the group 1 metals and is stable as H+. (c) [ReH9]2–? Names of ores? (a) Mg. AsH3. Al2O3 bauxite. and SiH4. Assign oxidation numbers to elements? H = –1. S = –2. (b) Al. S10. 10.7 9. 10. and (c) Pb.9? palladium hydride.9 are solids. 10. ice would be denser than water. 10.1. MgCO3 magnesite. PbS galena. and H3O+? The Lewis structures of these three species are: . 10. 9. (c) Hydrogen in group 14? There is no reason for hydrogen to be placed in this group.. especially in aqueous media. O = –2. GeH4 Reactions of hydrogen compounds? (a) Ca(s) + H2(g) → CaH2(s). Chemical characteristics of hydrides? (d) Lewis basicity? Ammonia. (b) KH? H = –1. Ionization energy increases across a period and decreases down a group. It also only has two electrons shared between two protons. 9.1 (b) Trends in ΔfGº? See Table 10. and HI are gases. S10. It has a high bond enthalpy. 10. NH3. Similarities? V (vanadium). none is a liquid.2.2 (c) Different molecular hydrides? Molecular hydrides are found in groups 13/III through 17/VII. Which of the following CH4. the strongest Bronsted acid. then P = +1. (b) Hydrogen in group 17? Hydrogen can fill its 1s orbital and make a hydride H–. SiH4. boiling points of HF. of LiNO3 decomposes in one step. 10. H3O+ should be trigonal pyramidal. AlH4–.13 Most likely to undergo radical reactions? (CH3)3SnH. 10. and GaH4–? Since AlH4– is more “hydride-like. (b) Increasing basicity toward a hard acid? H2Se < H2S < H2O.4 Sketch the thermodynamic cycle of Group 1 carbonate. (ii) protonation of a Brønsted base. Explain the differences in temperature of decomposition of LiNO3 and KNO3? KNO3 decomposes in two steps at two different temperatures. 10. 10. 10. and H2Se in order? (a) Increasing acidity? H2O < H2S < H2Se.1 Dihydrogen as an oxidizing agent? It’s reaction with an active s-block metal such as sodium. Only one resonance in the NMR at high temperature.15 The synthesis of binary hydrogen compounds? (i) direct combination of the elements. .20 Potential energy surfaces for hydrogen bonds? (See Figure 10.14 Arrange H2O.bond angles in period 2 hydrogen compounds reflect a greater degree of sp3 hybridization S11.2 Lattice enthalpies of formation? LiF is 625 kJ mol−1 and for NaF is 535 kJ mol−1. S11.17 Compare period 2 and period 3 hydrogen compounds? Period 2 compounds: . while the surface for the bifluoride ion has a single minimum. are all exoergic . 10.9) The surface for the H2O. and relatively low first ionization energies. or LiAlH4.tend to be weaker Brønsted acids and stronger Brønsted bases . M2CO3(s) CO2(g) 2M+(g) + CO32−(g) O2−(g) + CO2(g) S11. H2O. and NH3 are all higher than their respective period 3 homologues. H2S. 10.1 (a) Why are group 1 metals good reducing agents? They have one valence electron in the ns1 subshell.16 Compare BH4–.Several period 2 compounds exhibit strong hydrogen bonding. S11. CHAPTER 11 Self-tests S11. Cl– system has a double minimum. NaBH4. 2KNO2(s) → K2O(s) + 2NO2(g) + 1/2O2(g) Suggest a method for the preparation of BiH3? The redistribution BiH2Me.19 Describe the compound formed between water and Kr? A clathrate hydrate. LiNO3(s) → 1/2 Li2O(s) + NO2(g) + 1/4O2(g) methylbismuthine.21. the tin compound is the most likely to undergo radical reactions with alkyl halides.12 The reaction that will give the highest proportion of HD? Reaction (b) will produce 100% HD and no H2 or D2.18 M2O(s) + Predicted 7Li NMR of Li3N? Two peaks in the NMR spectrum at low temperature.6 3BiH2Me → 2BiH3 + BiMe3 10. S11. Exercises 11.ANSWERS TO SELF-TESTS AND EXERCISES 18 10.” it is the strongest reducing agent.5 .1 Change in cell parameter for CsCl? At 445 °C the CsCl structure changes to rock-salt and assumes the face centered cubic. 10. and (iii) metathesis using a compound such as LiH.except for B2H6. 2M+(g) + KNO3(s) → KNO2(s) + 1/2O2(g) .3 Trend is stability of Group 1 ozonides? Group 1 ozonides are less stable compared to the superoxides. 11.10 Predict the products of the following reactions? (a) CH3Br + Li → Li(CH3) + LiBr Calculate the lattice enthalpy of MgF2 and comment on how it will affect the solubility compared to MgCl2? MgF2 is 2991 kJ mol–1. .1 11. NaNH2 11.3 Synthesis of group 1 alkyls? Most alkyl lithiums are made using elemental lithium with the corresponding alkyl chlorides.2 (b) Why are group 1 metals poor They are large. 11. complexing agents? electropositive metals and have little tendency to act as Lewis acids. 8-coordinate Cs+. (b) Be or Sr. A = M(OH)2 M(OH)2 + CO2 → MCO3. CHAPTER 12 Trends of the fluorides and chlorides of the group 1 metals? Fluoride is a hard Lewis base and will form strong complexes with hard Lewis acids.2. C. 12. Exercises 12. form an acetate complex? Mg2+. D = MCl2. lower for CsF and LiI.2? Potassium ion.ANSWERS TO SELF-TESTS AND EXERCISES 19 11.4 Trends in solubility? Higher for LiF and CsI. C = MC2 MC2 + 2H2O → M(OH)2 + C2H2 M(OH)2 + 2HCl → MCl2 + 2H2O.6 should be close to ZnS-like structure. 12. B = MCO3 2MCO3 + 5C → 2MC2 + 3CO2. BaCl2 is ionic.5 Self-tests Why does beryllium fluoride form a glass when cooled from a melt? BeF2 adopts SiO2 like arrangement. Carbonates decompose to oxides. + NaOH ← H2O + Sodium metal + O2 → Na2O2 + heat → Na2O Predict whether (a) BeCl2 and (b) BaCl2 are predominantly ionic or covalent? BeCl2 is covalent. S12. 12. The trends reverse for the chloride ion.7 Thermal stability of hydrides versus decompose to carbonates? Hydrides elements.5 Why is magnesium hydroxide a much more effective antacid than calcium or barium hydroxide? Mg(OH)2 is sparingly soluble and mildly basic.2 Why are the properties of beryllium more similar to aluminium and zinc than to magnesium? Because of a diagonal relationship between Be and Al. 11. 11.3 Identify the compounds A. form a complex with C2. B. liquid (c) Li or K . + in S12.2 S12. 11.4 Which is more likely to lead to the desired result? (a) Cs+ or Mg2+. and D of the group 2 element M? M + H2O → M(OH)2.1 12.9 The effect of the alkyl group on the structure of lithium alkyls? Whether a molecule is monomeric or polymeric is based on the streric size of the alkyl group – less bulky alkyl groups lead to polymerization. Identify the compounds? + NH3 Calculate the lattice enthalpies for CaO and CaO2 and check that the above trend is confirmed? Calcium oxide and calcium peroxide are 3465 kJ mol–1 and 3040 kJ mol–1. will reduce solubility compared to MgCl2. 11.4 (b) MgCl2 + LiC2H5 → Mg(C2H5)Br + LiBr (c) C2H5Li + C6H6 → LiC6H5 + C2H6 Why are compounds of beryllium covalent whereas those of the other group 2 elements are predominantly ionic? Be has large polarizing power and a high charge density due.6 Use ionic radii to predict a structure type of BeSe? According to Table 3.8 The structures of CsCl and NaCl? 6-coordinate Na+. dissolve ammonia? Strontium.3 ↓ S12. N’’-trimethylB. (b) BCl3 and pyridine in hydrocarbon solution. Predict the number of lines and their relative intensities in the 1 H-NMR spectrum of BH4–? 4.38 to predict the nature of the bonding in BeBr .13 BH3NH3 + LiCl + H2 BCl3(g) + 3 EtOH(l) → B(OEt)3(l) + 3 HCl(g) BeTe. BaBr2 should be ionic 12. (c) BBr3 and F3BN(CH3)3? (b) BCl3 and pyridine in hydrocarbon solution? Predict structures for BeTe and BaTe.9 How do group 2 salts give rise to scaling from hard water? LiBH4 + NH4Cl S13.3 Use the data in Table 1.4.5 How many framework electron pairs are present in B H and to what structural category does it belong? Sketch its structure? 4 C2H5MgBr will be tetrahedral with two molecules of solvent coordinated to the magnesium. Group 1 hydroxides are more soluble than group 2 hydroxides.6 Explain why Group 1 hydroxides are much more corrosive to metals than Group 2 hydroxides? B nuclei have I = 3/2. arachno species. close to ZnS-like structure. BaTe.1 10 7.2 Which of the salts MgSeO or BaSeO would be expected to be more soluble in water? Write an equation for the reaction of LiBH4 with propene in ether solvent and a 1:1 stoichiometry and another equation for its reaction with ammonium chloride in THF with the same stoichiometry? MgSeO4 Simple alkenes are inert towards LiBH4. MgBr .7 and the Ketelaar triangle in Fig.8 4 4 Which Group 2 salts are used as drying agents and why? Anhydrous Mg. The structure of B4H10: Predict the products of the following reactions? (a) MgCl2 + 2LiC2H5 → 2LiCl + Mg(C2H5)2 (b) Mg + (C2H5)2Hg → Mg(C2H5)2 + Hg (c) Mg + C2H5HgCl → C2H5MgCl + Hg S13. 12. 2 Suggest a reaction or series of reactions for the preparation of N.B’’-trimethylborazine starting with methylamine and boron trichloride? Cl3BNCH3 + 3CH3MgBr (CH3)3B3N3(CH3)3 + 3Mg(Br. Relative intensity ratio is 1:3:3:1. close to CsCl-like structure. MgBr2 should be ionic.4 BeBr2 should be covalent.6-(CH ) C H MgBr dissolve in THF.12 The two Grignard compounds C H MgBr and 2. and BaBr . 12.10 11 S13.6 CHAPTER 13 Self-tests Propose a plausible product for the reaction between Li[B H ] and Al (CH ) ? 10 – [B10H11 (AlCH3)] 13 2 3 6 . 12. and Ca sulphates are preferred as drying agents.B’. 12.11 THF (a) BCl3 and ethanol? Salts of divalent ions have low solubility.6-(CH3)3C6 H2MgBr leads to a coordination number of two.7 12. Cl)2 5 2 S13. because of the higher affinity of Mg and Ca sulphates for water. 2 BCl3(g) + py(l) → Cl3B − py(s) 2 (c) BBr3 and F3BN(CH3)3? BBr3(l) + F3BN(CH3)3(s) → BF3(g) + Br3BN(CH3)3(s) S13. S13. What differences would be expected in the structures of the species formed in these solutions? 2 3 3 6 Write and justify balanced equations for plausible reactions between (a) BCl3 and ethanol.4.ANSWERS TO SELF-TESTS AND EXERCISES 20 12. N’. and therefore have higher OH− concentrations. The bulky organic group in 2. 12. 2. 4 BCl3 > BF3 > AlCl3 (a) BF3N(CH3)3 + BCl3 BCl3N(CH3)3 + BF3 (BCl3 > BF3) (b) BH3CO + BBr3 NR Thallium tribromide (1. B.3 Arrange the following in order of increasing Lewis acidity: BF3. AlCl3.ANSWERS TO SELF-TESTS AND EXERCISES 21 S13.11 g) reacts quantitatively with 0.172 kJ.B10C2H12 ⎯ ⎯→ 1. F2B– C2H4–BF2? 2BCl3 + 2Hg Æ B2Cl4 + 2HgCl2 B2Cl4 + 4AgF Æ B2F4 + 4AgCl B2F4 + C2H4 Æ F2CH2CH2BF2 Identify compounds A. dimers.7-B C H (Si(CH ) Cl) from 1. with reasons. TlBr3 + NaBr → NaTlBr4 13. Ether [HN(C2H5)3]Cl + NaBH4 H2 + H3BN(C2H5)3 + NaCl Draw the B12 unit that is a common motif of boron structures. devise a synthesis for the Lewis acid chelating agent. What would be the problem with diborane as a fuel? -73. take a viewpoint along a C2 axis? . and the boron containing product of combustion is a solid. a hydrocarbon of your choice.2 Describe the bonding in (a) BF3? Covalent.8 Predict the products from the hydroboration of (a) (CH3)2C=CH2.9 Diborane is extremely toxic.1 Give a balanced chemical equation and conditions for the recovery of boron? B2O3 + 3Mg → 2B + 3MgO ΔH < 0 13. (b) BH3CO + BBr3→? 13. The combustion reaction is B2H6 (g) + 3 O2(g) → 3 H2O (g) + B2O3 (s). (b) CH CH? (Me)2SalCl3 + GaBr3 → Me2SGaBr3 + AlCl3. and C? LiAlH4 (a) BH3 + (CH3)2C=CH2 B[CH2-CH(CH3)2]3 (b) BH3 + CH CH B(CH=CH2)3 Diborane has been used as a rocket propellant.257 g of NaBr to form a product A. (b) TlCl and formaldehyde (HCHO) in acidic aqueous solution? No reaction. it explodes in air.6 12 No.2B C H and other reagents of your choice? 10 10 2 2 10 3 2 2 13. write balanced chemical reactions (or no reaction) for (a) BF3N(CH3)3 + BCl3 →.12 C (a) A = B2H6 (b) B = B(OH)3 heat B Given NaBH4. Deduce the formula of A.7-B10C2H12 (90%) + 1. the chemical equation (or indicate no reaction) for reactions between (a) (CH3)2SAlCl3 and GaBr3? 13. Calculate the energy released from 1. (c) B2H6? Electron-deficient dimer.5 (c) C = B2O3 Does B2H6 survive in air? If not. give formulas and conditions for the synthesis of (a) B(C2H5)3. B2H6 + 3 O2 → B2O3 + 3 H2O2 Δ 1.11 A (b) 13. (b) Et3NBH3? (a) BCl3 + 3C2H5MgCl B(C2H5)3 + 3MgCl2 H2O CaF2 Using BCl3 as a starting material and other reagents of your choice. 13. (b) AlCl3? In the solid state. BCl3. Exercises 13. b) B4H10? a) 3 b) 2. At melting point. B2O3.00 kg of diborane given the following values of ΔfHө/kJ mol-1: B2H6 = 31.7 - Propose. 13. Identify the cation and anion? 13.2 .12B10C2H12 (10%) 1.7 Propose a synthesis for the polymer precursor 1. B2O3 = -1264.7 Predict how many different boron environments would be present in the proton-decoupled 11B-NMR of a) B5H11.8 → 1. write the equation for the reaction? Heat.7-B10C2H10Li2 + 2Si(CH3)2Cl2 B10C2H10(Si(CH3)2 + 2LiCl S13. and appropriate ancillary reagents and solvents. H2O = -242. a layered structure. In the light of this order. 13.10 BF3 13. or arachno.18 Heat (a) nido .14 13. Draw the structures of the products? (a) Ph3N3B3Cl3? 3 PhNH3+Cl− + 3 BCl3 → Ph3N3B3Cl3 + 9 HCl (b) Me3N3B3H3? 3 MeNH3+Cl− + 3 BCl3 → Me3N3B3Cl3 + 9 HCl Me3N3B3Cl3 + 3 LiH → Me3N3B3H3 + 3 LiCl The structures: . give the equations for the synthesis of [Fe(nido-B9C2H11)2]2-. 13.15 (a) Give a balanced chemical equation (including the state of each reactant and product) for the air oxidation of pentaborane(9).ANSWERS TO SELF-TESTS AND EXERCISES 22 + C2H2 (2) B10H12(SEt2)2 B10C2H12 + 2SEt2 + H2 (3) 2 B10C2H12 + 2EtO– + 4EtOH 2 B9C2H12– + 2B(OEt)3 + 2H2 (4) Na[B9C2H12] + NaH Na2[B9C2H11] + H2 THF (5) 2Na2[B9C2H11] + FeCl2 2NaCl + Na2[Fe(B9C2H11)2] 2- H B H B H 13.17 B H How many skeletal electrons are present in B5H9? 14 H H H H B B B B C C H H B (a) What are the similarities and differences in structure of layered BN and graphite (Section 13. (a) From its formula. (b) Their reactivity with Na and Br2? Graphite reacts. for the use of pentaborane as a fuel for an internal combustion engine? (a) 2B5H9 (l) + 12O2 (g) 5B2O3 (s) + 9H2O (l) 13. the number of cluster valence is the remainder of 44-20=24. nido. other than cost. (c) Verify by detailed accounting of valence electrons that the number of cluster valence electrons of B10H14 is the same as that determined in (b)? Starting with B10H14 and other reagents of your choice.9)? (b) Contrast their reactivity with Na and Br2. (b) Describe the probable disadvantages.19 Devise a synthesis for the borazines (a) Ph3N3B3Cl3 and (b) Me3N3B3H3.16 (b) The boron containing product of combustion is a solid. (b) Use Wade’s rules to determine the number of framework electron pairs for decaborane(14). (c) Suggest a rationalization for the differences in structure and reactivity. (b) 12. (c) Explain the differences? The large HOMO–LUMO gap in BN means it is more difficult to remove an electron from it than from the HOMO of graphite. starting with BCl3 and other reagents of your choice. boron nitride is unreactive. and sketch the structure of this species? (1) B10H14 + 2SEt2 B10H12(SEt2)2 + H2 B H H H H C C B H H B B Fe B H B B B H H B H B H 13. (c) The total number of valence elections is (10x3)+(14x1)=44. B2O3. B6H10 or B6H12? Give a generalization by which the thermal stability of a borane can be judged? B6H10 13. (a) Their structures? Both of these substances have layered structures. classify B10H14 as closo.13 H Which boron hydride would you expect to be more thermally stable. 13. B5H9. the acidity increases as the size of the borane increases. This also results in a material with a higher conductivity. is an arachno borane. (b) With bromine? Bromine can remove electrons from the π-symmetry HOMOs of graphite. and SiClF3. B10H14.4 Draw the structure and determine the charge on the cyclic anion [Si4O12]ν ? .1 Arrange the following boron hydrides in order of increasing Brønsted acidity. CHAPTER 14 + Self-tests S14.2 and above to calculate the standard enthalpy of formation of CH4 and SiH4? The anion is SiF5– S14. 14.20 Silicon forms the chlorofluorides SiCl3F. 14.2-B10C2H12 is a closo carborane 13. S14.1 CH3 Describe how the electronic structure of graphite is altered when it reacts with (a) potassium. (b) Account for the fact that the 19F NMR spectrum shows two fluorine environments? (a) The cation is [(CH3)4N]+ Give the structural type and describe the structures of B4H10. B5H9? In a series of boranes. (a) Use the VSEPR rules to determine the shape of the cation and anion in the product. The enthalpy of combustion of CH4 is −888 kJ mol−1 and the C–H and C–F bond enthalpies are −413 and −489 kJ mol−1 respectively? The bond enthalpy of a C–F bond is higher than the bond enthalpy of a C–H bond.2 Use the bond enthalpy data in Table 14.3 13 13 CO(g) + 2MnO2(s) → CO2(g) + Mn2O3(s) 2Li(s) + D2 → 2LiD(s) CO2(g) + LiD(et) → Li+D13CO2−(et) 13 - F F Si CH4: ΔfH= –61kJ mol–1 SiH4: ΔfH= +39 kJ mol–1 Propose a synthesis of D13CO2– starting from 13CO? CH3 H3C F F F (b) There are two different fluorine environments. 1.2B10C2H12? B4H10.3 SiF4 reacts with (CH3)4NF to form [(CH3)4N][SiF5]. Sketch the structures of these molecules? 14. and 1. is a nido borane.2 Explain why CH4 burns in air whereas CF4 does not.ANSWERS TO SELF-TESTS AND EXERCISES 23 Exercises 13.21 14. (b) bromine? N C H3 (a) With potassium? Potassium results in a material with a higher conductivity. and draw a structure for the probable structure of the deprotonated form of one of them: B2H6. B5H9. SiCl2F2. 10 Use data from Resource section 3 to determine the standard potential for each of the reactions in Exercise 14. metallic. 14. Na. a saline carbide. (D) RSiOSiR +H2O (E) SiR4 (F)? SiO2 YYYYY14. but +2 is the most stable oxidation state of Pb. Co. (b) Increase. draw a periodic table and indicate the elements that form saline. for the elements carbon (diamond) to tin (grey). In each case. elements Cr. Er. (b) Does the electrical conductivity of silicon increase or decrease when its temperature is changed from 20˚C to 40˚C? 14. Eg.8 Identify the compounds A to F:? (A) SiCl4 (B) SiRCl3 (C) RSi(OH)3.6 Predict the appearance of the 1H-NMR spectrum of Sn(CH3)4? Doublet (a) There is a decrease in band gap energy from carbon (diamond) to grey tin. H–Br = 366? ΔhH˚CCl4 = 110 kJmol–1 ΔhH˚CBr4 = 86 kJmol–1 14. Pm. and metalloid carbides? Ionic(silane) Metallic Metalloid carbides carbides carbides Group I Li.5 (b). V. Tb. Yb. Nb. Pb therefore displays the inert-pair effect. Ba Group 13 Al B elements Group 14 Si elements 3d-Block Sc. Ta. Ti. Eu. Tm. Bond enthalpies/(kJ mol–1): O–H = 463.14 Describe the preparation. Pr. (b) (i) Sn2+ + PbO2 + 4 H+ → Sn4+ + Pb2+ + 2H2O.2 and the additional bond enthalpy data given here to calculate the enthalpy of hydrolysis of CCl4 and CBr4. Charge = –8.9 (a) Summarize the trends in relative stabilities of the oxidation states of the elements of Group 14. Mn. Both reactions agree with the predictions made in Excercise 14.13 Preferably without consulting reference material. elements W. Ni 4d-Block Zr. Hf. write balanced chemical reactions or NR (for no reaction) for the following combinations. Re.08 V.31 V. Os 6d-Block Ac elements Lanthanides Ce. Fe.5. Ru 5d-Block La. Cs Group II Be. Mg. 14. Ca. Ho. and indicate the elements that display the inert pair effect. and explain how the answer fits the trends.7 Use the data in Table 14. structure and classification of (a) KC8. Lu 14. Sm. (i) Sn2+(aq) + PbO2(s) (excess) → (air excluded) (ii) Sn2+(aq) + O2(air) →? (a) +4 is the most stable oxidation state for the lighter elements. H–Cl = 431. comment on the agreement or disagreement with the qualitative assessment you gave for the reactions? (i) V = +1. 14. K. (ii) V= 1. elements Tc.ANSWERS TO SELF-TESTS AND EXERCISES 24 8- 14. (b) CaC2. There is a layered structure of alternating sp2 carbon atoms and potassium ions.11 O SiO2(s) + C(s) → Si(s) + CO2(g) ΔH < 0 GeO2(s) + 2H2(g) → Ge(s) + 2H2O(g) ΔH < 0 Si O O O Si Si O Give balanced chemical equations and conditions for the recovery of silicon and germanium from their ores? O O Si Sn-NMR (a) Describe the trend in band gap energy. (b) With this information in mind.12 O Predict the appearance of the spectrum of Sn(CH3)4? Doublet 119 14.5 14. (b) CaC2? Ca(l) + 2C(s) → CaC2(s) or CaO(s) + 3C(s) → CaC2(s) + CO(g) . Gd. (ii) 2Sn2+ + O2 + 4H+ → 2Sn4+ + 2H2O. Mo. (c) K3C60? (a) KC8? Formed by heating graphite with potassium vapor or by treating graphite with a solution of potassium in liquid ammonia. elements Sr. Dy. Nd. elements Rb. Indicate those elements that display the inert-pair effect? N (a) How many bridging O atoms are in the framework of a single sodalite cage? (b) Describe the (supercage) polyhedron at the centre of the Zeolite A structure in Fig.and two-dimensional structures. Exercises 15.3? (a) 48.17 Describe in general terms the nature of the [SiO3]n2ν ion in jadeite and the silicaalumina framework in kaolinite? The structures of jadeite and kaolinite consist of extended one. 15.6 cm3) = 4 strongly acidic OH groups per molecule. What is the chain length? 15. (c) Account for the large difference in costs between these two methods? 3 4 (a) high-purity phosphoric acid? 2Ca3(PO4)2 + 10C + 6SiO2 → P4− + 10CO + 6CaSiO3 P4 (pure) + 5O2 → P4O10 P4O10 + 6H2O → 4H3PO4 (pure) (b) Fertilizer grade H3PO4? Ca5(PO4)3OH + 5H2SO4 → 3H3PO4 (impure) + 5CaSO4 + H2O (c) Account for the difference in cost? .2. Are these reactions best described as electron-transfer processes or nucleophilic displacements? List the elements in Groups 15 and indicate the ones that are (a) diatomic gases. Self-tests S15. decide whether phosphorus or sulphur is likely to be the stronger oxidizing agent? Sulphur. Summarize the reactions that are used for the synthesis of hydrazine and hydroxylamine.1 P As Sb Bi O S Se Te (b) Eight sodalite cages are linked together to form the large cage of zeolite A.2 Type of element Nonmetal Nonmetal nonmetal metalloid metalloid nonmetal nonmetal nonmetal nonmetal Diatomic gas? yes Inert pair effect? No no No no no no yes no no no No No Yes No No No No (a) Give complete and balanced chemical equations for each step in the synthesis of H PO from hydroxyapatite to yield (a) high-purity phosphoric acid and (b) fertilizer-grade phosphoric acid. The [SiO32–]n ions in jadeite are a linear polymer of SiO4 tetrahedra. CHAPTER 15 When titrated against base a sample of polyphosphate gave end points at 30.4 From trends in the periodic table. S15. S15.3 S15. A molecule with 2 terminal OH groups and four further OH groups is a tetrapolyphosphate. S15.16 14. Is this puckered structure consistent with the VSEPR model? Yes.ANSWERS TO SELF-TESTS AND EXERCISES 25 Calcium carbide contains discrete C22– ions. The two-dimensional aluminosilicate layers in kaolinite represent another way of connecting SiO4 tetrahedra. What are the advantages of dimethylhydrazine over these fuels? Dimehylhydrazine ignites spontaneously. 14. (c) metalloids. (d) true metals. 14.6 cm3. (c) K3C60? A solution of C60 can be treated with elemental potassium.4 cm3)/(7. Na4SiO4(aq) + 4HCl(aq) → 4NaCl(aq) + SiO2(s) + 2H2O(l) 14. (b) nonmetals.4 and 45.1 Consider the Lewis structure of a segment of the structure of bismuth shown in Fig. and produces less CO2.15 Write balanced chemical equations for the reactions of K2CO3 with HCl(aq) and of Na4SiO4 with aqueous acid? K2CO3(aq) + 2HCl(aq) → 2KCl(aq) + CO2(g) + H2O(l) NH3 + ClO− + H+ → [H3N—Cl-O]− + H+ → H2NCl + H2O H2NCl + NH3 → H2NNH2 + HCl Both can be thought of as redox reactions. respectively.2 Refined hydrocarbons and liquid hydrogen are also used as rocket fuel. It is ionic.5 There are (30. is strikingly unreactive. (b) NO2– . Do either of the reactants differ from the usual practice of taking as reference state the most stable form of an element? 10 P4(s) + 5O2(g) → P4O10(s) 15. (c) reaction of the product from part (b) with a solution of CaCl and name the product? 4 2 (a) Oxidation of P4 with excess O2? P4 + 5O2 → P4O10 (b) Reaction of the product from part (a) with excess H2O? Write the balanced chemical equation corresponding to the standard enthalpy of formation of P O (s). 15.6 Compare and contrast the formulas and stabilities of the oxidation states of the common nitrogen chlorides with the phosphorus chlorides? 15.3 Ammonia can be prepared by (a) the hydrolysis of Li N or (b) the hightemperature. .10 4 The only isolable nitrogen chloride is NCl3. (d) N3–? 3 (a) Hydrolysis of Li3N? (a) HNO3? 6Li + N2 → 2Li3N 4NH3(aq) + 7O2(g) → 6H2O(g) + 4NO2(g) 2Li3N + 3H2O → 2NH3 + 3Li2O High temperature (b) Reduction of N2 by H2? (b) NO2–? N2 + 3H2 → 2NH3 2NO2(aq) + 2OH−(aq) → (c) Account for the difference in cost? The second process is considerably cheaper than the first. (c) AsCl ? 5 + (a) PCl4 ? tetrahedron. physical state (s.11 Without reference to the text.ANSWERS TO SELF-TESTS AND EXERCISES 26 Fertilizer-grade phosphoric acid involves a single synthetic step for a product that requires little or no purification. (b) PCl4–? A see-saw.5 2 3 at elevated temperatures: 2 NaN3 + 3NaOH + NH3(g) 2NaNH2(l) + N2O → NaN3 + NaOH + NH3 N2 itself. with a triple bond between the two atoms. 15. and H . and (ii) P(III) and P(V) are both about equally stable. (b) reaction of the product from part (a) with excess water. high-pressure reduction of N by H . Specify the structure. (c) NH OH. (c) Account for the lower cost of the second method? P4O10 + 6H2O → 4H3PO4 (c) Reaction of the product from part (b) with CaCl2? 2H3PO4(l) + 3CaCl2(aq) → 3 Ca3(PO4)2(s) + 6HCl(aq) 2 2 15. 15. (c) AsCl5? A trigonal bipyramid.7 Use the VSEPR model to predict the probable shapes of (a) PCl4+. 15. Both PCl3 and PCl5 are stable. (b) PCl4–. Give balanced chemical equations for each method starting with N .4 NO2−(aq) + NO3−(aq) + H2O(l) (c) NH2OH? cold aqueous acidic solution Show with an equation why aqueous solutions of NH NO are acidic? NO2−(aq) + 2HSO3−(aq) + H2O(l) → NH3OH+(aq) + 2SO42−(aq) NH4NO3(s) + H2O Æ NH4+ + NO3-(aq) (d) N3–? Carbon monoxide is a good ligand and is toxic. Why is the isoelectronic N molecule not toxic? 3NaNH2(l) + NaNO3 → 4 15. give the chemical equations and conditions for the synthesis of (a) HNO . or g). as appropriate. because lithium is very expensive. 15. Li. sketch the general form of the Frost diagrams for phosphorus (oxidation states 0 to +5) and bismuth (0 to +5) in acidic solution and discuss the relative stabilities of the +3 and +5 oxidation states of both elements? (i) Bi(III) is much more stable than Bi(V).9 2 2 Starting with NH3(g) and other reagents of your choice.8 Give balanced chemical equations for each of the following reactions: (a) oxidation of P with excess oxygen. l. and it is thermodynamically unstable. and allotrope of the reactants. C = NO 15. nitric oxide from an automobile exhaust. 15. (d) NH Cl? 5 3 4 (a) H2O? tetrahedral POCl3. Are HPO22− (c) D = N2O4 (d) E= NO2 (e) F = NH4+ .12 Are reactions of NO2– as an oxidizing agent generally faster or slower when pH is lowered? Give a mechanistic explanation for the pH dependence of NO2– oxidations? AS Cl2 When equal volumes of nitric oxide (NO) and air are mixed at atmospheric pressure a rapid reaction occurs. C.18 Identify the nitrogen compounds A. However. 15. B. Give an explanation for this observation in terms of the rate law and the probable mechanism? C (a) A = AsCl3 (b) B = AsCl5 (c) C = AsR3 (d) D = AsH3 2 2 4 15.16 Identify the compounds A. C. reacts slowly with air. Give balanced chemical equations for the reactions of the following reagents with PCl and indicate the structures of the products: (a) water (1:1). (b) water in excess. (c) AlCl .14 B 3RMgBr The rates of reactions in which nitrite ion is reduced are increased as the pH is lowered. to form NO and N O . HPO32– and H2PO2– ions will be much better reducing agents than oxidizing agents. and D? D LiAlH4 15. D.ANSWERS TO SELF-TESTS AND EXERCISES 27 and H2PO22− useful reducing agents? as oxidizing or Eº = 0. and E? (a) A = NO2 (b) B = HNO3. 15. which is present in the parts per million concentration range.15 Use standard potentials (Resource section 3) to calculate the standard potential of the reaction of H3PO2 with Cu2+.839 V.13 Cl2/hv A 2 The cis isomer gives two 19F signals and the trans isomer gives one signal. PCl5 + AlCl3 → [PCl4]+[AlCl4]− (d) NH4Cl? Cyclic molecules or linear chain polymers nPCl5 + nNH4Cl → −[(N = P(Cl)2)n]− + 4nHCl 15. PCl5 + H2O → POCl3 + 2HCl (b) H2O in excess? 2PCl5(g) + 8H2O(l) → 2H3PO4(aq) + 10HCl(aq) (c) AlCl3? tetrahedral.17 Sketch the two possible geometric isomers of the octahedral [AsF Cl ]– and explain how they could be distinguished by 19F– NMR? 4 The rate law must be more than first order in NO concentration. B. 2 Predict which oxidation states of Mn will be reduced by sulfite ions in basic conditions? (a) Formulas? H5TeO6– and HSO4– State whether the following oxides are acidic. Exercises 16. Use VSEPR theory to predict the shapes of the cation and anion? 4 3 3 4 SF4+ trigonal pyramidal. CO2. Both are trigonal pyramidal. Predict whether any of the following will be reduced by thiosulfate ions.068 V. 16.10 in acidic solution. S2O62−and S2O32−. calculate ΔrG for the disproportionation of hydrogen superoxide (HO ) into O and H O . and P. (b) Catalysis by Cr2+? of decomposing H2O2.ANSWERS TO SELF-TESTS AND EXERCISES 28 15. Cr2+ is not capable (c) The disproportionation of HO2? ΔrG° = 157 kJ. OH. (b) Offer a plausible explanation for this difference? (b) An explanation? Tellurium is a larger element than sulfur and can increase its coordination number. N2O. H4P2O6. Self-tests S16. SO .O3SO2SO32–? Mn (+VII.19 Use the Latimer diagrams in Resource section 3 to determine which species of N and P disproportionate in acid conditions? The species of N and P that disproportionate are N2O4. (b) K Te ? 2 2 CHAPTER 16 16.11 (a) The disproportionation of H2O2 and HO2? 1. and predict reactions with OH-.1 4 16.7 Probable structures of SO2F. +V. Fe3+.2 2 ethylenediamine is a better solvent than sulfur dioxide.51 HO2- 2 H2O2 16.1 3 S2O82– > SO42– > SO32– 2 The decomposition of H2O2 is thermodynamically favored in presence of Br– S16.70 g) reacts with SF (0. CO is neutral. (b) Is Cr a likely catalyst for the disproportionation of H O ? (c) Given the Latimer diagram 2 16. . BF4− tetrahedral. 16. Cu+. 16. MgO. (a) Write a balanced equation for 4 . 2 2 Use the standard potential data in Resource section 3 to predict which oxoanions of sulfur will disproportionate in acidic conditions? The SeO32– is marginally more stable in acid solutions. Al O .and (CH3) 3NSO2.6 In Presence of Cl– The decomposition is of H2O2 thermodynamically favored in presence of Cl– . +VI. +IV.SO32–. . or amphoteric: CO . Co3+? SF reacts with BF to form [SF ][BF ]. O or O—H .4 Which of the solvents ethylenediamine (which is basic and reducing) or SO (which is acidic and oxidizing) might not react with (a) Na S .5 Determine whether the decomposition of H O is spontaneous in the presence of either Br or Cl ? 2 2 2 16.S2O32–. For the disproportionation of H2O2 (part (a)).displaces F— or (CH3) 3N—. ΔrG° = 103 kJ. K O.8 3 (a) Use standard potentials (Resource section 3) to calculate the standard potential of the disproportionation of H O in acid solution. and Al2O3 are amphoteric. . P O . SO3. NH3OH+.81 g) to form an ionic product. and compare the result with its value for the disproportionation of H O ? 2 2 2 16. Use the standard potential data in Resource section 3 to predict whether SeO32– is more stable in acidic or basic solution? Fe3+ and Co3+ will be reduced. +III) will be reduced by sulfite ions in basic solution. ө 2– (a) Give the formula for Te(VI) in acidic aqueous solution and contrast it with the formula for S(VI). . in acidic conditions: VO2+. NO. basic. neutral. P2O5.3 Which hydrogen bond would be stronger: S—H . 16. CO? 2 Rank the following species from the strongest reducing agent to the strongest oxidizing agent: SO42–.9 2 2+ 2 O2 -0.13 1. 2 5 3 2 2 16. S? O–H hydrogen bonds are stronger.12 Tetramethylammonium fluoride (0. MgO and K2O are acidic. 2 Cl− + 2 H2O + electricity → Cl2 + H2 + 2 OH− For Br and I: Exercises 17. anode: 2 Cl−(aq) → Cl2(g) + 2 e− cathode: 2 H2O(l) + 2 e− → 2 OH−(aq) + H2(g) light yell ow yell ow- - unwanted reaction: 2 OH−(aq) + Cl2(aq) → ClO−(aq) + Cl−(aq) + H2O(l) . write out the halogens and noble gases as they appear in the periodic table.13 (a) SF4 + (CH3)4NF → [(CH3)4N]+[SF5]−.2 Both. (c) hardness of the halide ion. ICl2–.2 Predict the 19F-NMR pattern for IF7? For F: 2 resonances. and IF2–.3.1 Preferably without consulting reference material. (c) How many lines would be observed in the F-NMR spectrum of the anion? 19 16.0) 2 C l2 gas lower Har dnes s of hali de ion hard est soft er Col or Describe how the halogens are recovered from their naturally occurring halides and rationalize the approach in terms of standard potentials. The three centers contribute four electrons to three molecular orbitals. Give the chemical equation for the unwanted reaction that would occur if OH migrated through the membrane and into the anode compartment? – A drawing of the cell is shown in Figure 17. (b) electronegativity. C. NaIO . one nonbonding. (b) square pyramidal structure. Sketch a choralkali cell. C = S2N2. D = K2S2O3. and one antibonding. Identify the sulfur-containing compounds A. SO2 is cheaper. (d) color? F Phy sica l stat e Electron egativit y gas highest (4. one bonding. B = S4N4. Which of the reducing agents SO (aq) or Sn (aq) would seem practical from the standpoints of thermodynamic feasibility and plausible judgements about cost? Standard potentials are given in Resource section 3? B r2 liqu id lower soft er I2 soli d lowest soft est H e gas N e A r K r X e gas gas gas gas gree n dark redbro wn dark viol et Col orle ss colo rless colo rless colo rless colo rless 3 2+ 2 17.1 One source of iodine is sodium iodate. or g) at room temperature and pressure. l. Give balanced chemical equations and conditions where appropriate? 2 X− + Cl2 → X2 + 2 Cl− I−) 17.3 F2 + H2SO4 → CaSO4 + 2 HF From the perspective of structure and bonding. IBr2–. S17. and indicate the trends in (a) physical state (s.ANSWERS TO SELF-TESTS AND EXERCISES 29 the reaction and (b) sketch the structure of the anion. E = S2O62−. (c) two F environments. and F? A = S2Cl2. S17. indicate several polyhalides that are analogous to [py–I–py]+.3 (X− = Br−. E. D. and describe their bonding? 2 HF + 2 KF → 2 K+HF2− 2 K+HF2− + electricity → F2 + H2 + 2 KF For Cl: Examples include I3–. F = SO2. Show the half-cell reactions and indicate the direction of diffusion of the ions. CHAPTER 17 Self-tests S17. B. . 1 the mer isomer. FClO3 is pyramidal. (c) predict how many 19F-NMR signals would be observed in IF and X? Sketch the form of the vacant s* orbital of a dihalogen molecule and describe its role in the Lewis acidity of the dihalogens? 3 3 a). 1.9 Use the VSEPR model to predict the shapes of SbCl . MCl3F3. 2. 2. two. (b) Explain the difference in basicity? The strong electron-withdrawing effect in NF3 reduces the basicity. (a) identify X. (c) a plausible structure for trimethylsilyl cyanide? 2 2 (a) The reaction of NCCN with NaOH? NCCN(aq) + 2 OH−(aq) → CN−(aq) + NCO−(aq) + H2O(l) (b) The reaction of SCN– with MnO2 in aqueous acid? 2SCN−(aq) + MnO2(s) + 4 H+(aq) → (SCN)2(aq) + Mn2+(aq) + 2 H2O(l) (c) The structure of trimethylsilyl cyanide? Trimethylsilyl cyanide contains an Si–CN single bond. boils at –129˚C and is devoid of Lewis basicity.93 g of [(CH ) N]F to form a product X. the lower molar mass compound NH boils at –33˚C and is well known as a Lewis base. the cis isomer. c).6 The shape of anion IF4– is Square planar.84 g of IF reacts with 0. and [ClF ]+? 5 Based on the analogy between halogens and pseudohalogens write: (a) the balanced equation for the probable reaction of cyanogen. 17. (b) Describe the probable origins of the difference in basicity? 3 3 (a) Difference in volatility? Ammonia exhibits hydrogen bonding. (b) the equation for the probable reaction of excess thiocyanate with the oxidizing agent MnO (s) in acidic aqueous solution. 17. 2 Cl2 and F2. 17. By contrast.4 shapes of IF and the cation and anion in X. (a) Describe the origins of this very large difference in volatility. the fac isomer.IF3. Nitrogen trifluoride. b) IF3: X = IF4N(CH3)4 Different possible arrangements of Since the 2σu* antibonding orbital is the LUMO for a X2 molecule. FClO . Indicate how many fluorine environments would be indicated in the 19F-NMR spectrum of each isomer? 4 17. The shape of cation (CH3)4N+ is Tetrahedral. and ClF6 is octahedral. IF4–.ANSWERS TO SELF-TESTS AND EXERCISES 30 17.7 3 2 3 3 MCl4F2. (b) use the VSEPR model to predict the 3 3 4 6 SbCl5 is trigonal bipyramidal. with aqueous sodium hydroxide.11 Sketch all the isomers of the complexes MCl F and MCl F . 17. the trans isomer.5 Which dihalogens are thermodynamically capable of oxidizing H O to O ? 2 17. NF .8 Given that 1. 17.10 Indicate the product of the reaction between ClF and SbF ? 5 + 5 – [ClF4] [SbF6] . (CN) . one. it is the orbital that accepts the pair of electrons from a Lewis base. Write plausible Lewis structures for (a) ClO and (b) I O and predict their shapes and the associated point group? 2 (b) The preparation of [IF6] [SbF6]? 2 6 (a). 17. (a) SbF5? increases the acidity of BrF3. 3 2 (a) Describe the expected trend in the standard potential of an oxoanion in a solution with decreasing pH. is readily oxidized by strong oxidants.15 Predict whether each of the following solutes is likely to make liquid BrF a stronger Lewis acid or a stronger Lewis base: (a) SbF . IF7 + SbF5 → [IF6][SbF6] 17. The difference lies in iodine’s ability to expand its coordination shell. it will not form an explosive mixture with BrF3. Predict whether each of the following compounds is likely to be dangerously explosive in contact with BrF and explain your answer: (a) SbF . 17.788 V 17. IF7. (b) Demonstrate this phenomenon by calculating the reduction potential of ClO4– at pH = 7 and comparing it with the tabulated value at pH = 0? (a) The expected trend? E decreases as the pH increases.14 Predict the structure and identify the point group of ClO F? trigonal pyramidal. (b) SF .20 6 6 + 7 Explain why CsI (s) is stable with respect to the elements but NaI (s) is not? 3 3 Large cations stabilize large. V = 0. (c) F2? No.24 Which oxidizing agent reacts more readily in dilute aqueous solution. (b) CH OH.12 (a) Use the VSEPR model to predict the probable shapes of [IF ] and IF . 17. perchloric acid or periodic acid? Give a mechanistic explanation for the difference? Periodic acid. (b) Relative stabilities? Periodic acid is thermodynamically more stable.21 3 5 6 (a) The formulas are HBrO4 and H5IO6.17 . pentagonal bipyramid. 17. (b) Which is the more stable? 17. 2 (a) SbF5? Since SbF5 cannot be oxidized. (b) E at pH 0 and pH 7 for ClO4–? Eº = 1.22 + 5 17. 2 17. (b) Give a plausible chemical equation for the preparation of [IF ][SbF ]? 17. (c) F .18 (a) Give the formulas and the probable relative acidities of perbromic acid and periodic acid. (b) SF6? No effect on the acidity or basicity of BrF3. unstable anions.16 Predict the appearance of the F-NMR spectrum of IF ? Two resonances. (d) S Cl ? 3 5 2 3 2 (b) CH3OH? Methanol. (c) CsF? Increases the basicity of BrF3.ANSWERS TO SELF-TESTS AND EXERCISES 31 17. . 6 (a) The structures of [IF6]+ and IF7? octahedral. being an organic compound. Cs symmetry.13 Predict the shape of the doubly chlorine– bridged I Cl molecule by using the VSEPR model and assign the point group? 2 (b) I2O5? 6 A planar dimer (I2Cl6).23 (d) S2Cl2? S2Cl2 will be oxidized to higher valent sulfur fluorides. Write an equation (or NR for no reaction) for the interaction of [NR ][Br ] with I in CH Cl solution and give your reasoning? – 4 3 2 Br3− + I2 → 2 IBr + Br− 2 2 With regard to the general influence of pH on the standard potentials of oxoanions.201 V (see Appendix 2). (c) CsF? 17.19 IF6+. At pH 7. 19 17. explain why the disproportionation of an oxoanion is often promoted by low pH? Low pH results in a kinetic promotion: protonation of an oxo group aids oxygen– halogen bond scission. The formation of Br from a tetraalkylammonium bromide and Br is only slightly exoergic. 25 (a) For which of the following anions is disproportionation thermodynamically favourable in acidic solution: OCl . or photolyze Xe and F2 in glass: Xe(g) + F2(g) → XeF2(s) (a) NH4ClO4? Ammonium perchlorate is a dangerous compound. expensive inert By means of balanced chemical equations and a statement of conditions.3 Which of the following compounds present an explosion hazard? (a) NH ClO . (d) Co2+? No. ClO2– . (b) V3+.6 (a) Give a Lewis structure for XeF7–? Exercises 18. (b) XeF6? High temperature. Write a balanced equation for the decomposition of xenate ions in basic solution for the production of perxenate ions.4 Draw the Lewis structures of (a) XeOF4? (b) XeO2F2? (c) XeO64−? Use standard potentials to predict which of the following will be oxidized by ClO– ions in acidic conditions: (a) Cr3+. (c) Fe2+.ANSWERS TO SELF-TESTS AND EXERCISES 32 17. (b) Mg(ClO ) .1 Explain why helium is present in low concentration in the atmosphere even though it is the second most abundant element in the universe? Helium present in today’s atmosphere is the product of ongoing radioactive decay. (c) NaClO . Explain your reasoning? 4 4 2 4 2 4 6 4 2 least Argon. 17. (c) The atmosphere? 18. . 17. describe a suitable synthesis of (a) XeF2? Xe and F2 at 400ºC. 18. (b) V3+? Yes. 18. (c) XeO3? Xe(g) + 3 F2(g) → XeF6(s) XeF6(s) + 3 H2O(l) → XeO3(s) + 6 HF(g) (c) NaClO4? Not an explosion hazard.27 Which of the noble gases would you choose as (a) The lowest-temperature refrigerant? Helium. but have a large excess of F2: (b) Mg(ClO4)2? Not an explosion hazard.2 2 The rates of disproportionation are probably HClO > HClO2 > ClO3–. xenon. determine them from a table of standard potentials. (d) [Fe(H O) ][ClO ] .26 (b) An electric discharge light source requiring a safe gas with the lowest ionization energy? Xenon. (d) [Fe(H2O)6] [ClO4]2? An explosion hazard.) (b) For which of the favourable cases is the reaction very slow at room temperature? 18. and oxygen − − 4− 2 HXeO4 (aq) + 2 OH (aq) → XeO6 (aq) + Xe(g) + O2(g) + 2 H2O(l) Give the formula and describe the structure of a noble gas species that is isostructural with (a) ICl4−? XeF4 (c) BrO3–? Trigonal pyramidal geometry. 18.1 (b) IBr2–? Linear geometry. since Fe(II) can be oxidized to Fe(III). (b) Speculate on its possible structures by using the VSEPR model and analogy with other xenon fluoride anions? Pentagonal bipyramid. ClO2– . Isostructural with XeO3.5 CHAPTER 18 Self-tests S18. (c) Fe2+? Yes. (d) Co2+? (a) Cr3+? No. and ClO4– ? (If you do not know the properties of these ions. since the N atom of the NH4+ ion is in its lowest oxidation state and can be oxidized. (d) ClF? Isostructural with the cation XeF+. Isostructural with XeF2. 5. D. two other lines symmetrically distributed around the central line. The slipperiness of MoS2 is because of the ease with which one layer can glide over another. is more likely to form a sulfide in the presence of H S? (b) Rationalize your answer with the trends in hard and soft character across Period 4. Illustrate the trend using standard potentials in acidic solution for Groups 5 and 6. and those for which 2+ 2 Discrete molecular species such as Re3Cl123– or Re3Cl9(PPh3)3 are formed.4 For each part. 2 Cr3+ + 3 H2MoO4 + 2 H2O (c) MnO4– (aq) + Cr3+ (aq) →? 6 MnO4− + 10Cr3+ + 11H2O → 6Mn2+ + 5Cr2O72− + 22 H+ 19. VO2 S19. C. Exercises Without reference to a periodic table. 1:3:3:1 quartet 18. Ni (aq) or Mn (aq).8 Identify the xenon compounds A. (b) indicate the metals that form difluorides with the rutile or fluorite structures. and (c) indicate the region of the periodic table in which metal-metal bonded halide compounds are formed.7 Use molecular orbital theory to calculate the bond order of the diatomic species E2+ with E=He and Ne? He22+ = 0. Strong central line.6 Preferably without reference to the text (a) write out the d block of the periodic table. sketch the first series of the d block. 19.3 State the trend in the stability of the group oxidation state on descending a group of metallic elements in the d block. those for which the group oxidation number can be reached but is a powerful oxidizing agent by O.2 Explain why the enthalpy of sublimation of Re(s) is significantly greater than that of Mn(s)? As atoms become heavier. Indicate those elements for which the group oxidation number is common by C.10 Predict the appearance of the 19F-NMR spectrum of XeOF4. B. The group oxidation number increases in stability as you descend a group. Predict the appearance of the 129Xe-NMR spectrum of XeOF3+.ANSWERS TO SELF-TESTS AND EXERCISES 33 18.5 Describe the probable structure of the compound formed when Re3Cl9 is dissolved in a solvent containing PPh3? 2+ More likely for Ni2+ to form a sulphide. CHAPTER 19 Self-tests S19. 19. more energy is needed to vaporize them. including the symbols of the elements.3 Suggest a use for molybdenum(IV) sulfide that makes use of its solid-state structure.2 S19. give balanced chemical equations or NR (for no reaction) and rationalize your answer in terms of trends in oxidation states. and E? A = XeF2(g) − B = [XeF]+[MeBF3] C = XeF6 D = XeO3 E = XeF4 (g).1 Refer to the appropriate Latimer diagram in Resource section 3 and identify the oxidation state and formula of the species that is thermodynamically favoured when an acidic aqueous solution of V2+ is exposed to oxygen? (a) Cr2+ (aq) + Fe3+ (aq) →? Cr2+(aq) + Fe3+(aq) → Cr3+(aq) + Fe2+(aq) (b) CrO42– (aq) + MoO2 (s) →? 2 CrO42− + 3 MoO2(s) + 10 H+ → + +4. giving one example? . 19.1 (a) Which ion. (c) Give a balanced chemical equation for the reaction. Ni2+(aq) + H2S(aq) → NiS(s) + 2 H+(aq) 19. 18. Rationalize your suggestion? MoS2 used as a lubricant. 18.5.9 the group oxidation number is not achieved by N? 19. Hardness decreases from left to right in the d block. Ne22+ = 0. 9 (a) [Re(O)2 (py)4]+. molybdenum-molybdenum quadruple bond. and d bonding and antibonding orbitals.ANSWERS TO SELF-TESTS AND EXERCISES 34 Metal–metal bonded halide compounds are found within bold border. ability to accept electron density from an OH– or O2– ligand increases. and state the basis for your answer: (a) MoO42– (aq) plus Fe2+ (aq) in acidic solution? No reaction.7 Write a balanced chemical equation for the reaction that occurs when cis[RuLCl(OH2)]+ (see Fig. 19. NbCl . and WCl ? Rationalize the differences and speculate on the structures? (a) NiI2? ionic compound with significant degree of covalent character 2 4 2 2 (b) NbCl4? significantly covalent.2 V is made strongly basic at the same potential. noting any features of interest? (c) ReCl5 (s) plus KMnO4(aq)? 5 ReCl5(s) + 2 MnO4−(aq) + 12 H2O(l) → 5 ReO4−(aq) + 2 Mn2+(aq) + 25 Cl−(aq) + 24 H+(aq) (d) MoCl2(s) plus warm HBr(aq)? 6 MoCl2(s) + 2 Br−(aq) → [Mo6Cl12]2−(aq) + Br2(aq) (e) TiO(s) with HCl(aq) under an inert 2TiO (s) + 6H+ (aq) Æ atmosphere? 3+ 2Ti + H2(g) + 2H2O(l) E˚ = +0.11 As oxidation state of the metal increases. p. Give other examples and a reason for the redox state of the metal center affecting the extent of protonation of coordinated oxygen.10 Which of the following are likely to have structures that are typical of (a) predominantly ionic. Give plausible balanced chemical reactions (or NR for no reaction) for the following combinations. Write a balanced equation for each of the successive reactions when this same complex at pH = 6 and +0. (d) [VOCl4]2–? The first three are dioxo complexes and will have cisdioxometal structural units. (c) metal-metal bonded compounds: NiI . (f) Cd(s) added to Hg2+(aq)? Hg2+ (aq) + Cd(s) Æ Hg(l) + Cd2+ (aq) Indicate the probable occupancy of s. no metal–metal bond in this molecule. 19. cis-[RuIILCl(OH2)]+ + H2O → cis[RuIIILCl(OH)]+ + H3O+ + e− (e) WCl2 ? metal–metal bonding. cis-[RuIIILCl(OH2)]+ + H2O → cis[RuIVLCl(OH)]+ + H3O+ + e− 19.13 Addition of sodium ethanoate to aqueous solutions of Cr(II) gives a red diamagnetic product.9) in acidic solution at +0. (b) significantly covalent. (d) VOCl42–. measured at 25oC? Higher oxidation states become more stable on descending a group.0 V. FeF . PtS. and the bond order for the following tetragonal prismatic complexes? . 19. 19.12 Explain the differences in the following redox couples.2 V is exposed to progressively more oxidizing environments up to +1. Draw the structure of the product. (b) [Cr2(O2CC2H5)4]? σ2π4δ2. (b) [V (O)2 (ox)2]3–. (b) The preparation of [Mo6O19]2−(aq) from K2MoO4(s)? 6 MoO42−(aq) + 10 H+(aq) → [Mo6O19]2−(aq) + 5 H2O(l) (c) [Cu2(O2CCH3)4]? σ2π4δ2δ*2π*4σ*2 configuration. 19.chromium– chromium quadruple bond.37 V. (c) FeF2? ionic. has a square-pyramidal structure with an apical oxo ligand. − cis-[Ru LCl(OH2)] + Ox + OH → cis[RuIIILCl(OH)]+ + Red + H2O (d) PtS ? significant amount of covalent character. 19. Example: Sc5Cl6. (c) [Mo(O)2(CN)4]4–.8 Speculate on the structures of the following species and present bonding models to justify your answers: (a) [Mo2(O2CCH3)4]? σ2π4δ2 configuration. II + 19. 800 cm–1 are of this type. four unpaired electrons.1 Determine the configuration (in the form t2g x egy or ext2y.: 2. Assign these transitions using the d3 Tanabe-Sugano diagram and selection rule considerations. MF2 lattice enthalpies would increase from Mn(II) to Zn(II).700 cm–1 and 23.and lowspin d7 configurations? A high-spin d7 configuration is t25geg2. a band at 23 800 cm-1 with εmax= 5130 dm3 mol-1 cm-1. S20. (b) [Fe(OH2)6]2+? d6. where relevant. as appropriate). NiF2 (3060 kJ mol–1). the bond order of 2.400 cm–1 is probably a charge transfer band. 0.0 should also be expected [0. S20.4Δ0. [Ru2 (CH3COCH3)2(ClCO2)4]: 2+. Using the bonding scheme depicted in Figure 19.8 Δo. and a very strong band at 32 400 cm-1.6 Identifying ground terms (Hint: Because d9 is one electron short of a closed shell with L = 0 and S = 0. Suggest an interpretation of the photoelectron spectra of [Fe(C5H5)2] and [Mg(C5H5)2] shown in Fig. LFSE = 0. high spin. low spin. What is its electron 3 2 configuration? t2g eg S20.5 +.8 Use the same Tanabe-Sugano diagram to predict the energy of the first two spinallowed quartet bands in the spectrum of [Cr(OH2)6]3+ for which Δo = 17 600 cm-1 and B = 700 cm-1.5 What terms arise from a p1d1 configuration? 1F and 3F terms are possible.7 What terms in a d2 complex of symmetry correlate with the 3F and terms of the free atom? An F terms 3 T1g. .14 O O S20. probably 2Eg ← 4A2g.18? In the spectrum of [Mo(CO)6]. S20. in line with the energy level scheme [0. (a) [Co(NH3)6]3+? d6. 3T2g.4Δ0.3 S20. Exercises 20. If it were not for ligand field stabilization energy (LFSE).5 (8-3)]. What is the LFSE for both high. LFSE = 2.4 Account for the variation in lattice enthalpy of the solid fluorides in which each metal ion is surrounded by an octahedral array of F– ions: MnF– (2780 kJ mol–1). (b) 3d9? 2D (called a “doublet D” term). no unpaired electrons. Similarly. 17500 cm–1 and 22400 cm–1.000 cm–1 is a clue that it is a spin-forbidden transition. CoF2 (2976 kJ mol–1).8.has lowlying π* orbitals.has a very weak band near 16 000 cm-1. a band at 17 700 cm-1 with εmax= 5160 dm3 mol-1 cm-1. FeF2 (2926 kJ mol–1). the ionization energy around 8 eV was attributed to the t2g electrons that are largely metal-based.) O O Consider the two ruthenium complexes in Table 19.9 The spectrum of [Cr(NCS)6]3.) (a) 2p2? 3P (called a “triplet P” term). High spin.2 The magnetic moment of the complex [Mn(NCS)6]4– is 6. The (iii) The band at 32. so it is likely that the bands at 17.1 (i) The very low intensity of the band at 16. CHAPTER 20 Oh 1 D are are Self-tests S20. and 3A2g. treat it on the same footing as a d1 configuration. (Hint: NCS. 20. and the ligand-field stabilization energy in terms of ΔO or ΔT and P for each of the following complexes using the spectrochemical series to decide. [Ru2Cl2(ClCO2)4].ANSWERS TO SELF-TESTS AND EXERCISES 35 differences in the 6–8 eV region can be attributed to the lack of d electrons for Mg(II).06μB. confirm the bonding orders and electron configurations given in the table. (ii) Spin-allowed but Laporte-forbidden bands typically have ε ~ 100 M–1 cm–1. O O O Cr Cr O 19.5x(84)]. since its intensity is too high to be a ligand field (d–d) band.19. which are likely to be high-spin and which low-spin. and ZnF2 (2985 kJ mol–1). the number of unpaired electrons. mixed valence at the Ru center. S20. D terms 1 T2g and 1Eg. S20. high spin.3 20. identify the one that has the larger LFSE: (a) [Cr(OH2)6]2+ or [Mn(OH2)6]2+ (b) [Mn(OH2)6]2+ or [Fe(OH2)6]3+ (c) [Fe(OH2)6]3+ or [Fe(CN)6]3(d) [Fe(CN)6]3– or [Ru(CN)6] (e) tetrahedral [FeCl4]2–or tetrahedral [CoCl4]2– (a) The chromium complex.11 Identify the ground term from each set of terms: (a) 3F. [CoCl4]2– is blue. Spin-only contributions are: complex N μSO = [(N)(N + 2)]1/2 3+ [Co(NH3)6] 0 0 [Fe(OH2)6]2+ 4 4. the “single” electronic transition is really the superposition of two transitions. S = 5/2? 6S (b) L = 3. low spin. assign each color to one of the complexes. by possessing ligands that are π-acids or by possessing ligands that are strong σ-bases (or both). CoO (3988). 4P. For each of the following pairs of complexes. (e) [W(CO)6]? d6. The band shown in Fig.2Δ0. (g) Tetrahedral [Ni(CO)4]? d10. S = 3/2? 4F (c) L = 2. of the following values of oxide lattice enthalpies (in kJ mol–1). (b) 5D. high in the spectrochemical series. 3P.ANSWERS TO SELF-TESTS AND EXERCISES 36 (c) [Fe(CN)6]3–? d5. LFSE = 0. and (ii) LFSE. 1I? 5D (c) 6S.9 [Ni(CO)4] 0 0 Solutions of the complexes [Co(NH3)6]2+.are colored.6 ΔT.9 [Fe(CN)6]3– 1 1. and so the excited state possesses eg degeneracy. 20. Thus there are two ways for a complex to develop a large value of Δ0. LFSE = 0. S = 1? 3P 20. the complex turns violet and is high-spin with two unpaired electrons. one unpaired electron. 20. TiO (3878). NiO (4071)? There are two factors that lead to the values: (i) decreasing ionic radius from left to right across the d block. one from an Oh ground-state ion to an Oh excitedstate ion.9 [W(CO)6] 0 0 [FeCl4]2– 4 4.S): (f) Tetrahedral [FeCl4]2−? d6. (a) L = 0. 20. One is pink. 20. (e) The Co2+ complex.7 [Cr(NH3)6]3+ 3 3.5 Estimate the spin-only contribution to the magnetic moment for each complex in Exercise 20. Therefore.1. [Co(OH2)6]2+ (both Oh). low spin.9 The spectrum of d1 Ti3+(aq) is attributed to a single electronic transition eg ← t2g. . (c) [Fe(CN)6]3– (d) [Cr(NH3)6]3+? d3. 1G? 3F.and P(C6H5)3 are ligands of similar field strength. and [CoCl4]2. S = 1/2? 2D (d) L = 1.2 Both H. and a lower energy transition from an Oh ground-state ion to a lower energy distorted excited-state ion. [Co(NH3)6]2+ is yellow.8 Bearing in mind the Jahn-Teller theorem.10 Write the Russell–Saunders term symbols for states with the angular momentum quantum numbers (L. and the third is blue. LFSE = 1. Interpret the change in terms of structure? Shift from square planar to tetragonal complex. 3H. Considering the spectrochemical series and the relative magnitudes of ΔT and ΔO. 20. 2I? 6S . 1P.3 is not symmetrical and suggests that more than one state is involved. four unpaired electrons. (d) The ruthenium complex. 20.6 Interpret the variation. Recalling that phosphines act as π acceptors. VO (3913). low spin. All the compounds have the rocksalt structure: CaO (3460). is π-acceptor character required for strong-field behaviour? What orbital factors account for the strength of each ligand? No. SCN–.7 A neutral macrocyclic ligand with four donor atoms produces a red diamagnetic low-spin d8 complex of Ni(II) if the anion is the weakly coordinating perchlorate ion. [Co(OH2)6]2+ is pink. 0 unpaired electrons. LFSE is 2. including the overall trend across the 3d series. 3+ (b) The Fe complex. 20.4 20. another is yellow. three unpaired electrons. FeO (3921).20. MnO (3810). no unpaired electrons.4Δ0. Suggest how to explain this observation using the Jahn-Teller theorem? The electronic excited state of Ti(OH2)63+ has the configuration t2g0eg1. 1G. LFSE = 2. Predict the structure of [Cr(OH2)6]2+? Elongated octahedron.0Δ0. 3P. 4G. When perchlorate is replaced by two thiocyanate ions. 20. 20. CrO42–. How should these transitions be assigned? The first two transitions listed above correspond to two low-spin bands.21 Solutions of [Cr(OH2)6]3+ ions are pale blue– green but the chromate ion.20 Ordinary bottle glass appears nearly colorless when viewed through the wall of the bottle but green when viewed from the end so that the light has a long path through the glass. The very weak band in the red corresponds to a spinforbidden transition. The blue-green color of the Cr3+ ions in [Cr(H2O)6]3+ is caused by spinallowed but Laporte-forbidden ligand field transitions.8B. the complex is expected to be colorless. but in this case a single spin-allowed transition makes the complex colored and gives it a oneband spectrum.33 cm–1 and C = 3167. 20. What do you suggest for the origins of these transitions? First. The relatively low molar absorption coefficient is the reason that the intensity of the color is weak. No spin-allowed transitions are possible for the Fe3+. (c) High-spin [Fe(H2O)6]3+? 6A1g. The 1D and 3P terms lie.and 615 cm-1 in [Co(NH3)6]3+. such as [CoCl(NH3)5] –. E(3P) = A + 7B. The gas-phase ion V3+ has a 3F ground term. The very weak peak is most likely a spinforbidden ligand field transition. and a strong band at higher energy with εmax= 2 x 104 dm3 mol–1 cm–1.ANSWERS TO SELF-TESTS AND EXERCISES 37 20. the intense yellow color is due to LMCT transitions.is colorless whereas [CoF6]3– is colored but exhibits only a single band in the visible. E(1D) = A .18 The Racah parameter B is 460 cm–1 in [Co(CN)6]3.19 An approximately ‘octahedral’ complex of Co(III) with ammine and chloro ligands gives two bands with εmax between 60 and 80 dm3 mol–1 cm–1. one weak peak with .7 cm–1. (a) Which orbitals will be displaced from their position in the octahedral molecular orbital diagram by π interactions with the lone pairs of the Cl– ligand? (b) Which orbital will move because the Cl– ligand is not as strong a base as NH3? (c) Sketch the qualitative molecular orbital diagram for the C4v complex. The d6 Co3+ ion in [CoF6]3– is also high spin. but cyanide is also a πacid. 10 642 and 12 920 cm–1 above it. respectively.17 Explain why [FeF6]3. The color is associated with the presence of Fe3+ in the silicate matrix. is an intense yellow. The energies of the terms are given in terms of Racah parameters as E(3F) = A . 20. (b) [Ni(NH3)6]2+ (absorptions at 10750. which is only observed when looking through a long pathlength of bottle glass. 17500 and 28200 cm-1)? Δ0 = 10.22 Classify the symmetry type of the d orbital in a tetragonal C4v symmetry complex. 20. Consider the nature of bonding with the two ligands and explain the difference in nephelauxetic effect? Ammonia and cyanide ion are both σ-bases. 20. Suggest which transitions are responsible for the color? The faint green color. Give the Russell-Saunders terms of the configurations and identify the ground term? (a) 4 s1? 2S. The Cl atom lone pairs of electrons can form π molecular orbitals with dxz and dyz. 15400 and 26000 cm-1) ? Δ0 = 8500 cm–1 and B ≈ 770 cm–1. The ground term is 3P.13 εmax52 dm3 mol-1 cm-1.16 The spectrum of [Co(NH3)6]3+ has a very weak band in the red and two moderate intensity bands in the visible to near-UV. Characterize the origins of the transitions and explain the relative intensities. is caused by spin-forbidden ligand field transitions. Calculate the values of B and C for V3+. and so they will be raised in energy. (b) [Ti(H2O)6]3+? 2T2g.750 cm–1 and B ≈ 720 cm–1. 20. The oxidation state of chromium in dichromate dianion is Cr(VI). B = (12920 cm– 1 )/(15) = 861. 20. estimate ΔO and B for (a) [Ni(H2O)6]2+ (absorptions at 8500. 20. These metal atomic orbitals are π-antibonding MOs. 3P.3B + 2C.14 Write the d-orbital configurations and use the Tanabe–Sugano diagrams (Resource section 6) to identify the ground term of (a) Low-spin [Rh(NH3)6]3+? 1A1g. 1S. the intense band at relatively high energy is undoubtedly a spin-allowed charge-transfer transition. 1 D. The two bands with εmax = 60 and 80 M–1 cm–1 are probably spin-allowed ligand field transitions. where the Cl lies on the z-axis.15 Using the Tanabe-Sugano diagrams in Resource section 6.12 (b) 3p2? 20. 21. 21. 20. The volume of activation is found to be 111. why may it be difficult to characterize an aqua ion as labile or inert? The rate of an associative process depends on the identity of the entering ligand and. N3–. k = 1. period 5 and 6 d-block metals have stronger metal ligand bonds.2 Given the reactants PPh .1 The rate constants for the formation of [CoX(NH3)5]2+ from [Co(NH3)5OH2]3+ for X = Cl2.6 A Pt(II) complex of tetramethyldiethylenetriamine is attacked by Cl.3 cm3 mol–1.105 times less rapidly than the diethylenetriamine analogue. explain how to estimate ΔT from an assignment of the two chargetransfer transitions.are 18 500 and 32 200 cm–1. Therefore.2 × 102 s–1. What is the mechanism of the substitution? Dissociative.4 Write the rate law for formation of [MnX(OH2)5]1 from the aqua ion and X–. . 20. 2- Use the data in Table 21. Given that the wavenumbers of the two transitions in MnO4. The greater steric hindrance.24 The lowest energy band in the spectrum of [Fe(OH2)6]3+ (in 1M HClO4) occurs at lower energy than the equivalent transition in the spectrum of [Mn(OH2)6]2+.8 to estimate an appropriate value for KE and calculate kr2 for the reactions of V(II) with Cl. and [PtCl ] .1 M−1s−1 S21. is just equal to ΔT.[PtCl2(NH3)(PPh3)]? S21.23 Consider the molecular orbital diagram for a tetrahedral complex (based on Fig.8 The pressure dependence of the replacement of chlorobenzene (PhCl) by piperidine in the complex [W(CO)4(PPh3)(PhCl)] has been studied. Account for this observation on the basis of a dissociative rate-determining step. The oxidation state of manganese in permanganate anion is Mn(VII).3 3 3 4 Metal centers with high oxidation numbers have stronger bonds to ligands than metal centers with low oxidation numbers. therefore. The extra charge of the iron complexes keeps the eg and t2g levels close Exercises 21. no ligand field transitions are possible. How would you undertake to determine if the reaction is d or a? rate = (kKE[Mn(OH2)62+][X−])/(1 + KE[X−]) 21. Furthermore.ANSWERS TO SELF-TESTS AND EXERCISES 38 20. Is the reaction d or a? d. 21. Octahedral complexes of metal centers with high oxidation numbers or of d metals of the second and third series are less labile than those of low oxidation number and d metals of the first series of the block. for which npt = 3. What does this value suggest about the mechanism? Mechanism must be dissociative. 21.71 = 5. NH .3 The reactions of Ni(CO)4 in which phosphines or phosphites replace CO to give Ni(CO)3L all occur at the same rate regardless of which phosphine or phosphite is being used. 21. Propose efficient routes to both cis. 21. KE = 1 M–1.2x102 dm3 mol–1 s–1. even though ΔT cannot be observed directly.5 CHAPTER 21 Self-tests S21.if the observed second-order rate constant is 1. E(t2) – E(e) = 13700 cm–1.1 Calculate the second-order rate constant for the reaction of trans-[PtCl(CH3) (PEt3)2] with NO2– in MeOH. The difference in energy between the two transitions. from [W(CO)4L(PhCl)] increases with increase in the cone angle of L.22 ? k2(NO2−) = 100.and trans. Br2. which is d0.2 If a substitution process is associative. Explain why this is. PhCl. and SCN¯ differ by no more than a factor of two. Explain this observation in terms of an associative ratedetermining step. it is not an inherent property of [M(OH2)6]n+.7) and the relevant d-orbital configuration and show that the purple color of MnO4ions cannot arise from a ligand-field transition. What does this observation suggest about the mechanism is mechanism? The dissociative.7 The rate of loss of chlorobenzene. 9 21. What experimental data might be used to distinguish between the two pathways? The inner-sphere pathway: [Co(N3)(NH3)5]2+ + [V(OH2)6]2+ → [[Co(N3)(NH3)5]2+. (iii) The implication is that a complex without protic ligands will not undergo anomalously fast OH– ion substitution.15 A two-step synthesis for cis.and outer-sphere pathways for reduction of azidopentaamminecobalt(III) ion with V2+(aq). followed by the coordination of 1. [Ni(OH2)6]2+? The order of increasing rate is [Ir(NH3)6]3+ < [Rh(NH3)6]3+ < [Co(NH3)6]3+ < [Ni(OH2)6]2+ < [Mn(OH2)6]2+. 21. There is a kinetic pathway with activation parameters Δ‡ H = 1101 kJ mol–1 and Δ ‡S = 142 J K –1 mol –1. (b) Changing the leaving group from Cl– to I–? The rate decreases. 21. [Ir(NH3)6]3+. (d) Increasing the positive charge on the complex? Rate increase. A possible mechanism is loss of one dimethyl sulfide ligand.16 State the effect on the rate of dissociatively activated reactions of Rh(III) complexes of each of (a) an increase in the positive charge on the complex? decreased rate.10 Does the fact that [Ni(CN)5]3– can be isolated help to explain why substitution reactions of [Ni(CN)4]2– are very rapid? For a detectable amount of [Ni(CN)5]3– to build up in solution.10-phenanthroline. that of base hydrolysis. 21.14 Reactions of [Pt(Ph) 2(SMe2) 2] with the bidentate ligand 1.12 How does each of the following affect the rate of square-planar substitution reactions? (a) Changing a trans ligand from H to Cl? Rate decreases. [V(OH2)5]2+. [Mn(OH2)6]2+.13 The rate of attack on Co(III) by an entering group Y is nearly independent of Y with the spectacular exception of the rapid reaction with OH–. (c) Adding a bulky substituent to a cis ligand? The rate decreases. [V(OH2)6]2+} → {[Co(N3)(NH3)5]2+. [V(OH2)6]2+} {[Co(N3)(NH3)5]2+.17 Write out the inner. [V(OH2)5]2+. 21. (b) changing the leaving group from NO3– to Cl–? Decreased rate. Predict the products of the following reactions: (a) [Pt(PR3)4]2+ + 2Cl–? cis[PtCl2(PR3)2]. H2O} {[Co(N3)(NH3)5]2+. Propose a mechanism. (c) cis-[Pt(NH3)2(py)2]2+ + 2Cl–? trans[PtCl2(NH3)(py)].and trans[PtCl2(NO2) (NH3)] – start with [PtCl4]2–? Put in order of increasing rate of substitution by H2O the complexes: [Co(NH3)6]3+. (c) changing the entering group from Cl– to I–? This change will have little or no effect on the rate. 21. H2O} → [(NH3)5Co–N=N=N–V(OH2)5]4+ [(NH3)5Co–N=N=N–V(OH2)5]4+ → [(NH3)5Co– N=N=N–V(OH2)5]4+ [(NH3)5Co–N=N=N–V(OH2)5]4+ → [Co(OH2)6]2+ + [V(N3)(OH2)5]2+ The outer sphere pathway: [Co(N3)(NH3)5]2+ + [V(OH2)6]2+ → {[Co(N3)(NH3)5]2+ [V(OH2)6]2+} .11 (ii) The anomalously high rate of substitution by OH– signals an alternate path. What is the implication of your explanation for the behaviour of a complex lacking Brønsted acidity on the ligands? (i) The general trend: octahedral Co(III) complexes undergo dissociatively activated ligand substitution. the forward rate constant kf must be numerically close to or greater than the reverse rate constant kr.10-phenanthroline (phen) give [Pt(Ph) 2phen]. 21. [Rh(NH3)6]3+. 21.ANSWERS TO SELF-TESTS AND EXERCISES 39 21. Explain the anomaly. (b) [PtCl4]2– + 2PR3? trans-[PtCl2(PR3)2]. (d) changing the cis ligands from NH3 to H2O? Decreased rate. total = 18. Na and CH3I.4. K12 = e [nF ε /RT] where εo = 0.6 × o 103 dm3mol−1s−1. What product and quantum yield do you predict for substitution of [W(CO)5(py)] in the presence of excess triethylamine? Is this reaction expected to be initiated from the ligand field or MLCT excited state of the The product will be complex? [W(CO)5(NEt3)].22 From the spectrum of [CrCl(NH3)s]2+ shown in Fig.31 Jmol−1K−1 and T = 298 K. (b) k11 (Cr3+/Cr2+) = 1 × 10−5 dm3mol−1s−1.41V) = 1.4.2 What is the electron count for and oxidation number of platinum in the anion of Zeise’s salt. (b) [Co(NH3)6]3+ (EO(Co3+/Co2+) = +0.6 Propose a synthesis Mn(CO)4(PPh3)(COCH3) starting [Mn2(CO)10].[V(OH2)6]2+} → {[Co(N3)(NH3)5]+ . propose a wavelength for photoinitiation of reduction of Cr(III) to Cr(II) accompanied by oxidation of a ligand. and the quantum yield will be 0. CHAPTER 22 Self-tests No. 21.18 The compound [Fe(SCN)(OH2)5]2+ can be detected in the reaction of [Co(NCS)(NH3)5]2+ with Fe2+(aq) to give Fe3+(aq) and Co2+(aq). Substitution of these values in the MarcusCross relationship gives k12 = 8. n = 1. f12 = 1.1 o dm3mol−1s−1. 2 2 S22.4 Which of the two iron compounds Fe(CO)5 and [Fe(CO)4(PEt3)] will have the higher CO stretching frequency? Which will have the longer M–C bond? Fe(CO)5 S22.3 What is the formal name of [Ir(Br)2(CH3)(CO)(PPh3)2]? Dibromocarbonylmethylbis(triphenylphosphine)iridium(III). (a) [(η6-C7H8)Mo(CO)3] (49)? η6-C7H8 = 6 electrons. R = 8. What does this observation suggest about the mechanism? Appears to be an inner-sphere electron transfer reaction.26 V). Comment on the relative sizes of the rate constants. f12 = 1.41V) = 0.32. S22. (c) k11 (Cr3+/Cr2+) = 1 × 10−5 dm3mol−1s−1. (b) [(η7-C7H7)Mo(CO)3]+(51)? η7-C7H7+ = 6 electrons. R = 8. F = 96485 C.53 × 103 dm3mol−1s−1 (b) [Co(NH3)6] 3+ −2 3 −1 −1 k = 1.41 V) and each of the oxidants [Ru(NH3)6]3+ (EO(Ru3+/Ru2+) = +0. n = 1. Substitution of these values in the MarcusCross relationship gives k12 = 2. Comment on the relative sizes of the rate constants (a) k11 (Cr3+/Cr2+) = 1 × 10−5 dm3mol−1s−1. PPh3.5 Show that both are 18-electron species. 21.19 × 107 dm3mol−1s−1. n = 1.41 × 10 dm mol s Relative sizes? The reduction of the Ru complex is more thermodynamically favoured and faster.67 V.26 × 1019. k22 (Ru3+/Ru2+ for the hexamine complex) = 6. F = 96485 C.07 V). Using these values we get K12 = 9. A flash photolysis study reveals a spectrum that can be assigned to the intermediate W(CO)5.41V) = 1. carbonyl = 2 electrons. Using these values gives K12 = 1.255 V) and the oxidants (a) [Ru(NH3)6]3+ (EO(Ru3+/Ru2+) = + 0.51 × 1015 The dm3mol−1s−1.21. S22. K12 = e[nF ε /RT] where εo = 1. Mo atom = 6.32 × 108. Substitution of these values in the MarcusCross relationship gives k12 = 3. {[Co(N3)(NH3)5] +.95 × 103 dm3mol−1s−1.18 V. S22. K12 = e[nF ε /RT] where εo = 0. +4. Electron count. [PtCl3(C2H4)]–? Treat CH =CH as a neutral two-electron donor. for with .ANSWERS TO SELF-TESTS AND EXERCISES 40 {[Co(N3)(NH3)5]2+.81 × 1028. F = 96485 C.31 Jmol−1K−1 and T = 298 K.07 V – (–0.07 V). 16.26 V – (–0. ~250 nm.77 V) and [Ru(bpy)3]3+ (EO(Ru3+/Ru2+) = +1. Mo atom = 6. [V(OH2)6]3+} Using these values gives K12 = 1.48 V. total = 18. the quantum yield is approximately 0. k22 (Fe3+/Fe2+ for the aqua complex) = 1.31 Jmol−1K−1 and T = 298 K.20 Calculate the rate constants for electron transfer in the oxidation of [Cr(OH2)6]2+ (EO–(Cr3+/Cr2+) = –0. f12 = 1. [Fe(OH2)6]3+ (EO(Fe3+/Fe2+) = +0.19 Calculate the rate constants for electron transfer in the oxidation of [V(OH2)6]2+ (Eσ (V3+/V2+) = –0. R = 8. In the presence of excess phosphine. oxidation number.21 photochemical substitution of [W(CO)5(py)] (py = pyridine) with triphenylphosphine gives W(CO)5(P(C6H5)3).77 V – (−0.10 V). k22 (Ru3+/Ru2+ for the bipy complex) = 4 × 108 o dm3mol−1s−1. (a) [Ru(NH3)6]3+ k = 4.1 Is Mo(CO)7 likely to be stable? S22.[V(OH2)6]3+} → {[Co(OH2)6]2+ . 20. [V(OH2)6]3+ 21. 3 COs = 6. 21. (j) Co(η5-C5H5)(η4-C4Ph4)? cyclopentadienyltetraphenylcylcobutadinecob alt(I). 18e–. S22.2 Sketch an η2 the interactions of 1. 18e–. By using the same molecular orbital diagram. PPh3 Pd Pd + PhCl Cl Cl The decrease in both coordination number and oxidation number by 2.1 Mn(CH3)(CO)5 + PPh3 → Mn(CO)4(PPh3)(COCH3) S22. PMe3 would be preferred. 18e–. 17e– (d) [Fe(CO)4]2–? tetracarbonylferrate(–2). (k) [Fe(η5-C5H5)(CO2)]–? dicarbonylcyclopentadienylferrate(0). (b) Mn2(CO)10? decacarbonyldimanganese(0). propose a structure for [Fe4(Cp)4(CO)4]. draw the structures of.11 Assess the relative substitutional reactivities of indenyl and fluorenyl (86) compounds? Fluorenyl compounds are more reactive than indenyl.16e(i) Pd(Cl)(Me)(PPh3)2? chloromethylbis(triphenylphosphine) palladium(II). (l) Cr(η6-C6H6)(η6-C7H8)? benzenecycloheptatrienechromium(0).8 S 22. whereas [Pt(PEt3)2 (Me)(Cl)] The ethyl group in does not? [Pt(PEt3)2(Et)(Cl)] is prone to β-hydride elimination. (e) La(η5-Cp*)3? tris(pentamethylcyclopentadienyl)lanthanum(I II). Name the species. comment on whether the removal of an electron from [Fe(η5-Cp)2] to produce [Fe(η5-Cp)2]+ should produce a substantial change in M–C bond length relative to neutral ferrocene. The 1H NMR spectrum is a single line even at low temperatures. It will not. 18e–.) Bridging (a) Fe(CO)5? Pentacarbonyliron(0). (g) Fe(CO)4(PEt3)? tetracarbonyltriethylphosphineiron(0). or both? (Substitution of η5-C5H5 ligands for CO ligands leads to small shifts in the CO stretching frequencies for a terminal CO ligand. 18e–. From this spectroscopic information and the CVE. S22.9 The IR spectrum of [Ni2(η5-Cp)2(CO)2] has a pair of CO stretching bands at 1857 cm–1 (strong) and 1897 cm–1 (weak). (h) Rh(CO)2(Me)(PPh3)? dicarbonylmethyltriphenylphosphinerhodium( I). and give valence electron counts to the metal atoms? Do any of the complexes deviate from the 18-electron rule? If so. TA tetrahedron with 4 terminal Cp rings and four capping COs. 18e–.7 S22. how is this reflected in their structure or chemical properties? 22. The compound [Fe4(Cp)4(CO)4] is a darkgreen solid. (c) V(CO)6? hexcarbonylvanadium(0).ANSWERS TO SELF-TESTS AND EXERCISES 41 Mn2(CO)10 + 2 Na → 2Na[Mn(CO)5] Exercises Na[Mn(CO)5] + CH3I → Mn(CH3)(CO)5 + NaI 22. 16e–.10 If Mo(CO)3L3 is desired. 18e–. Cl Ph3P Cl Cl Ph PPh3 Ph3P (n) Ni(η5-C5H5)NO? cyclopentadineylnitrosylnickel(0). 18e–. Does this complex contain bridging or terminal CO ligands.12 Show that the reaction is an example of reductive elimination? (m) Ta(η5-C5H5)2Cl3? trichlorobiscyclopentadineyltanatalum(V). 18e–.13 Explain why [Pt(PEt3)2(Et)(Cl)] readily decomposes. which of the ligands P(CH3)3 or P(t-Bu)3 would be preferred? Give reasons for your choice. S22.4butadiene with a metal atom and (b) do the same for an η4 interaction. 22. .18e–. Its IR spectrum shows a single CO stretch at 1640 cm–1.3 What hapticities are possible for the interaction of each of the following ligands S22. (f) Fe(η3-allyl)(CO)3Cl? allyltricarbonylchloroiron(II). and other reagents of your choice. Without consulting reference material. Justify your answer. Cp* is a stronger donor ligand than Cp. given iron metal. η3. very common for group 9 and group 10 elements. Cs symmetry complex. Is the selection of method based on thermodynamic or kinetic considerations? (1) Mo(s) + 6 CO(g) → Mo(CO)6(s) (high temperature and pressure required) (2) 2 CoCO3(s) + 2 H2(g) + 8 CO(g) Co2(CO)8(s) + 2 CO2 + 2 H2O (b) Does the electron count for each metal in your structure agree with the 18electron rule? If not. (c) C6H6? η6.4 spectrum? Check your answer and give the number of expected bands for each by consulting Table 22.8 Provide plausible reasons for the differences in IR wavenumbers between each of the following pairs: (a) Mo(PF3)3(CO)3 2040. 22. (c) Co(η3-C3H5)(CO)3? 18. or η1.5 State the two common methods for the preparation of simple metal carbonyls and illustrate your answer with chemical equations. (a) Ni(η3-C3H5)2 16.6 Suggest a sequence of reactions for the preparation of Fe(CO)3(dppe). The IR data indicate that all C5H5 ligands are pentahapto and probably in identical environments. Fe(s) + 5 CO(g) → Fe(CO)5(l) (high temperature and pressure required) Fe(CO)5(l) + diphos(s) → Fe(CO)3(diphos)(s) + 2 CO(g) 22. and Cs. 1939 cm–1 vs. The Ir complex. PMe3 is primarily a σ-donor ligand. is nickel in a region of the periodic table where deviations from the 18-electron rule are common? No. 22. PF3 is primarily a π-acid ligand. propose a structure. 1991 cm–1 versus Mo(PMe3)3(CO)3 1945. If the electron count deviates from 18.11 Which metal carbonyl in each of (a) [Fe(CO)4]2– or [Co(CO)4]– (b) [Mn(CO)5]– or [Re(CO)5]–should be the most basic toward a proton? What are the trends on which your answer is based ? (a) [Fe(CO)4]2– The trend involved is the greater affinity for a cation that a species with a higher negative charge has. → The reason that the second method is preferred is kinetic. Deviations from the rule are common for cyclopentadienyl complexes to the right of the d block. 1928 cm–1? CO bands of the Cp* complex are lower in frequency than the corresponding bands of the Cp complex. (b) Co(η4-C4H4)(η5-C5H5)? 18. η2.10 Decide which of the two complexes W(CO)6 or IrCl(CO)(PPh3)2 should undergo the faster exchange with 13CO. is the deviation explicable in terms of periodic terms.ANSWERS TO SELF-TESTS AND EXERCISES 42 with a single d-block metal atom such as cobalt? (a) C2H4? η2 (b) Cyclopentadienyl? Can be η5. η4. and η2. 3 Draw plausible structures and give the electron count of (a) Ni(η3-C3H5)2 (b) (c) Co(η3Co(η4-C4H4)(η5-C5H5) C3H5)(CO)3. 22.9 (d) Cyclooctadiene? η2 and η4 . which of these should display the greatest number of CO stretching bands in the IR 22.7 Suppose that you are given a series of metal tricarbonyl compounds having the respective symmetries C2v. MnCp*(CO)3 2017. It has an A mechanism. 22. dppe (Ph2PCH2CH2PPh2). (a) On the basis of these data. CO. η6. η4. 22. D3h. The compound Ni3(C5H5)3(CO)2 has a single CO stretching absorption at 1761 cm–1. (e) Cyclooctatetraene?η8. . 1851 cm–1? CO bands of the trimethylphosphine complex are 100 cm–1 or more lower in frequency. (b) MnCp(CO)3 2023. 22.7. both starting with Mn2(CO)10.17 (a) Reflux Mo(CO)6 with cycloheptatriene to give [Mo(η6-C7H8)(CO)3]. indicate the probable number of carbonyl ligands in (a) W(η6-C6H6)(CO)n. However. C. react with MeI to give MnMe(CO)5. (a) 3 (b) 2 (c) 12 22. (ii).20 Which compound would you expect to be more stable. treatment of TiCl4 at low temperature with MeLi or LiCH2SiMe3 gives compounds that are stable at room temperature. K+PF6–. (i) Reduce Mn2(CO)10 with Na to give Mn(CO)5–. Use this information and the 18-electron rule to identify the compounds A. and (c) Ru3(CO)n. Compound D can be treated with bromine to yield E or with Na/Hg to give compound F. E. B. with one using Na and one using Br2? You may use other reagents of your choice. which has the molecular formula W(C3H5)(C5H5)(CO)3.16 W CO CO C: B: η3 -2 C O -C O -H Fe OC Fe CO CO A OC H CO B [π-C5H5Fe(CO)2]2 C The compound B shows two 1H NMR resonances due to Fe-H proton and the aromatic Cp ring. The compounds formed are TiR4. treat with the trityl tetrafluoroborate to give [Mo(η7C7H7)(CO)3]BF4. Rationalize these observations. 22. The trend involved is the greater M–H bond enthalpy for a period 6 metal ion relative to a period 4 metal ion in the same group.18 W CO CO CO η1 22. 22. When Fe(CO)5 is refluxed with cyclopentadiene compound A is formed which has the empirical formula C8H6O3Fe and a complicated 1H NMR spectrum. 22. D = C5H5Mo(CO)3. results in the formation of a salt. A.19 W CO CO CO η2 Treatment of TiCl4 at low temperature with EtMgBr gives a compound that is unstable above 270ºC. Compound C has a single 1H-NMR resonance and the empirical formula C7H5O2Fe. F = C5H5Mo(CO)3Na. + F e (C O ) 5 + A: Suggest syntheses of (a) [Mo(η7C7H7)(CO)3]BF4 from Mo(CO)6 and (b) from [IrCl2(COMe)(CO)(PPh3)2] [IrCl(CO)(PPh3)2]? When Mo(CO)6 is refluxed with cyclopentadiene compound D is formed which has the empirical formula C8H5O3Mo and an absorption in the IR spectrum at 1960 cm–1. There are absorptions in the IR spectra of E and F at 2090 and 1860 cm–1. and C all have 18 valence electrons: identify them and explain the observed spectroscopic data. E = C5H5Mo(CO)3Br.15 Na[W(η5-C5H5)(CO)3] reacts with 3chloroprop-1-ene to give a solid.12 Using the 18-electron rule as a guide. and C. Compound C has the molecular formula [W(C3H6)(C5H5)(CO)3]PF6. Treating compound A with hydrogen chloride and then potassium hexafluorophosphate. 22. B. 22. Compound A loses carbon monoxide on exposure to light and forms compound B. ethyl has the low energy β-hydride elimination decomposition. Sketch a structure for each.RhCp2 or RuCp2? Give a plausible explanation for the difference in terms of simple bonding concepts . to give MnBr(CO)5.ANSWERS TO SELF-TESTS AND EXERCISES 43 (b) The rhenium complex. and F all have 18 valence electrons: identify them and explain the observed spectroscopic data. paying particular attention to the hapticity of the hydrocarbon. respectively.14 Give the probable structure of the product obtained when Mo(CO)6 is allowed to react first with LiPh and then with the strong carbocation reagent. Compounds D. displace the bromide with MeLi to give MnMe(CO)5.13 Propose two syntheses for MnMe(CO)5. 22. (b) React [IrCl(CO)(PPh3)2] with MeCl. Subsequent heating of B results in the loss of H2 and the formation of compound C. Oxidize with Br2. then expose to CO atmosphere to give [IrCl2(COMe)(CO)(PPh3)2]. Compounds A. which has the formula W(C3H5)(C5H5)(CO)2. Mo(CO)5(C(OCH3)Ph) 22. (b) Rh(η5C5H5)(CO)n. C shows a single 1H NMR resonance because of aromatic Cp ring. Compound A readily loses CO to give compound B with two 1H-NMR resonances. one at negative chemical shift (relative intensity one) and one at around 5ppm (relative intensity 5). CH3OSO2CF3. 22. Co4(CO)12 or Ir4(CO)12? Suggest an explanation. S23.3 Use the Frost diagrams and data in Resource section 2 to determine the most stable uranium ion in acid aqueous solution in the presence of air and give its formula. N(CH3)2. Exercises 23. The cobalt complex probably contains a trigonal-prismatic Co6 array. The symmetry-adapted orbitals of the two eclipsed C5H5 rings in a metallocene are shown in Resource Section 5. as does s. the dz2 orbital on the metal has a1′ symmetry. If the proposed mechanism is applicable. CHAPTER 23 Self-tests S23. choose the groups that might replace the group in boldface in (a) Co3(CO)9CH→ OCH3. and state how many a1’ molecular orbitals may be formed. for the reactions: (a) [Mn(CO)5(CF2)]+ + H2O → [Mn(CO)6]+ + 2HF? (i) The F atoms render the C atom subject to nucleophilic attack (ii)two equivalents of HF are eliminated (b) Rh(C2H5) (CO) RhH(CO)(PR3)2 + C2H4? elimination reaction.22 22.21 (b) Can these CVE values be derived from the 18-electron rule? No. what rate constant can be extracted from rate data? Rate = ka[RMn(CO)5] (a) What cluster valence electron (CVE) count is characteristic of octahedral and trigonal prismatic complexes? octahedral M6. giving your reasoning. 22. 90. (a) Fe(η5 – C5H5)2 + CH3COCl → 5 Fe(η – C5H5)(η5 – C5H4COCH3) + HCl 22.ANSWERS TO SELF-TESTS AND EXERCISES 44 RuCp2 has 18 electrons.27 (b) Cl Fe 22. p. Identify the s. (b) (OC)5MnMn(CO)5 → I. or CCH3? I. which would you expect to undergo the fastest exchange with added13CO. In the latter compound the H atom is attached to the Fe atom. CH2. The compound Ni(η5-C5H5)2 readily adds one molecule of HF to yield [Ni(η5C5H5)(η4-C5H6)]+ whereas Fe(η5-C5H5)2 reacts with strong acid to yield [Fe(η5C5H5)2H]+.26 Based on isolobal analogies.25 22. trigonal prismatic M6. Protonation of FeCp2 at iron does not change its number of valence electrons. (c) Determine the probable geometry of and [Co6(C)(CO)15]2[Fe6(C)(CO)16]2– ? The iron complex probably contains an octahedral Fe6 array. and it is postulated that these occur by initial breaking of an M-M bond. or SiCH3? SiCH3.1 (a) Give a balanced equation for the reaction of any of the lanthanoids with aqueous acid. and d orbitals of a metal atom lying between the rings that may have nonzero overlap. Three a1′ MOs will be formed. (b) Justify your answer with redox potentials and with a generalization on the most stable positive oxidation states for the lanthanoids. Give the equation for a workable reaction for the conversion of Fe(η5-C5H5)2 to Fe(η5C5H5) (η5-C5H4COCH3) and (b) Fe(η5C5H5) (η5-C5H4CO2H) 22. Provide a reasonable explanation for this difference. (PR3)2 → A β-hydrogen Given mechanism of CO insertion.1 Derive the ground state of the Tm3+ ion 3 H6. Co. thereby providing an open coordination site for the incoming ligand.24 + O O Cl A lC l 3 C H 2C l2 Cl Fe H 2 O . t-B u O K DME C O 2H Fe Sketch the a1’ symmetry-adapted orbitals for the two eclipsed C5H5 ligands stacked together with D5h symmetry. This is because metal–metal bond strengths increase down a group in the d block. S23. Write a plausible mechanism. (c) Name the two lanthanoids that have the greatest tendency to deviate from the usual positive oxidation . Suggest a strategy to ensure that the hydride is monomeric.2 The product of the reaction above is in fact a hydride bridged dimer (9). A simple approach would be to use the Cp rings substituted by bulky groups and examine rate of formation. UO22+ if sufficient oxygen is present.28 Ligand substitution reactions on metal clusters are often found to occur by associative mechanisms. 86.23 22. 2 Why does increased pressure reduce the conductivity of K+ more than that of Na+ in β-alumina? Because K+ is larger than Na+.6 Products of reactions? (a) LiNiO2.5 Synthesis of? (a) MgCr2O4 .heat (NH4)2Mg(CrO4)2 gradually to 1100-1200°C. high temperature.2 Explain the variation in the ionic radii between La3+ and Lu3+.9 Account for the similar electronic spectra of Eu3+ complexes with various ligands and the variation of the electronic spectra of Am3+ complexes as the ligand is varied. 24. 23.72 μB 23.10 Predict a structure type for BkN based on the ionic radii r(Bk3+) = 96 pm and r(N3−) = 146 pm.1 Synthesis for Sr2MoO4? Sealed tube.8 Suggest a synthesis of neptunocene from NpCl4? Np(COT)2 was prepared by the reaction of K2COT with NpCl4 under inert atmosphere. speculate on why Ce and Eu were the easiest lanthanoids to isolate before the development of ion-exchange chromatography.35. (b) SrFeO3Cl . The lanthanide contraction. The rock-salt structure. varies as a function of ligand. (c) 90Sr. How would you expect the first and second ionization energies of the lanthanoids to vary across the series? Sketch the graph that you would get if you plotted the third ionization energy of the lanthanoids versus atomic number? Identify elements at any peaks or troughs and suggest a reason for their occurrence? First and second IEs would show general increase across the lanthanoids. 23. 23. and decide which of the following highly radioactive nuclides are likely to present the greatest radiation hazard in the spent fuel from nuclear power reactors: (a) 39Ar. anomalies arise. S24.99 V for europium to 2. 23. With the third. 90Sr and 144Ce. The normal sites for cations in this structure are the tetrahedral holes.3 Rationalize the observation that FeCr2O4 is a normal spinel? The A2+ ions (Fe2+ in this example) occupy tetrahedral sites and the B3+ ions (Cr3+) occupy octahedral sites. 6 SrO(s) + Mo(s) + 2 MoO3(s) Æ 3 Sr2MoO4.6 Predict the magnetic moment of a compound containing the Tb3+ ion. 23. 24. unusual oxidation states were used in separation procedures. ? (a) Balanced equation 2 Ln(s) + 6 H3O+(aq) → 2 Ln3+(aq) + 3 H2(g) + 6 H2O(l) (b) Redox potentials The potentials for the Ln0/Ln3+ oxidations in acid solution range from a low of 1. and the splitting of the 5f subshell.2 What is a crystallographic shear plane? Both. CHAPTER 24 S24. 24.1 NiO doped with Li2O? The electronic conductivity of the solid increases owing to formation of Ni1–xLix)O. as well as the color of the complex.4 Wurtzite crystal structure and bottleneck? The wurtzite structure is shown in Figure 3. Eu2+. (b) Sr2WMnO6.4 7 23. 23. 24. S24. The bottleneck involves the space formed by three close packed anions.3 How might you distinguish between a solid solution and a series of discrete crystallographic shear plane structures? A solid solution would contain a random collection of crystallographic shear planes. . (d) 144Ce. (c) Ta3N5 . Self-tests Exercises 24.ANSWERS TO SELF-TESTS AND EXERCISES 45 state and correlate this deviation with electronic structure. Ce4+ and Eu2+. (b) 228Th. (c) Two unusual lanthanides Ce4+.3 From a knowledge of their chemical properties.5 Derive the ground state of the Tb3+ ion. 23.heat SrO + SrCl2 + Fe2O3 in a sealed tube.7 Explain why stable and readily isolable carbonyl complexes are unknown for the lanthanoids? Carbonyl compounds need back-bonding from metal orbitals of the appropriate symmetry.38 V. The 5f orbitals of the actinide ions interact strongly with ligand orbitals.11 Describe the general nature of the distribution of the elements formed in the thermal neutron fission of 235U. F6 23. 24. μ= 9.heat Ta2O5 under NH3 at 700°C. 25.14 × 106 nm2 (a factor of 104). 25. In copper aluminate spinel blue.10 Using LFSEs determine site preference for A= Ni(II) and B = Fe(III)? Better ligand field stabilization and a strong preference for inverse spinel.5 (a) Top-down versus bottom-up? The “topdown” approach requires one to “carve out” nanoscale features from a larger object.1 (a) Surface areas? 3. are superconductors. . 25. and P. 24. the site is tetrahedral. In Na3C60. (b) Advangtages and disadvantages? Topdown methods allow for precise control over the spatial relationships.9 Magnetic measurements on ferrite? inverse spinel. 25. 24. B. the spins on different metal centers are coupled into an antiparallel alignment.12 Classify oxides as glass-forming or nonglass-forming? BeO. 24.16(a).21 nm and the diameter is 8. (ZnP2)O6.7 %. 24. Transition metal and rare earth oxides are typically nonglassforming.4 Band energies for QD versus bulk semiconductor? The energies of the band edges for a QD nanocrystal are more widely separated.14 × 102 nm2 versus 3.13 Which metal sulfides might be glass forming? Metalloid and nonmetal sulfides.20 24. P.7 Where might intercalated Na+ ions reside in the ReO3 structure? The unit cell for ReO3 is shown in Figure 24.16 Oxotetrahedral species in structures? Be. (b) Nanoparticles based on size? The 10 nm particle. and to some extent GeO2. all of the tetrahedral holes and all of the octahedral holes are filled with sodium cations. Exercises 25. which contains only Cu2+.8 Antiferromagnet ordering? In an antiferromagnetic substance.18 Mass percent of hydrogen in NaBH4 and hydrogen storage? 10. 24. and Zn replacing Si? (AlP)O4. S25.14 Examples of spinels containing? (a) (b) fluoride? sulfide? Zn(II)Cr(III)2S4 Li2NiF4 24. 24. Bottom-up methods framework Mg1-x(M)xH2y (M=Al.3 Why are QDs better for bioimaging? One light source can be used to excite different quantum dots. 24. (c) Nanoparticles based on properties? A nanoparticle should exhibit properties different from those of a molecule or an extended solid. 24. B2O3.22 Fulleride structures? In Na2C60. all of the tetrahedral holes are filled with sodium cations. S25.17 Formulas for structures isomorphous with SiO2 containing Al. or through electrochemical insertion of Li ions. the net magnetic moment goes to zero.ANSWERS TO SELF-TESTS AND EXERCISES 46 24. 24.15 Synthesis of LiTaS2? Direct reaction of TaS2 with BuLi.11 High-temperature superconductors? All except Gd2Ba2Ti2Cu2O11.2 Hemispherical imprint for 2 nm nanoparticle? The radius is 4.21 Color intensity differences in Egyptian blue pale versue blue-green spinel? The Cu site in Egyptian blue is square planar. 24.19 Formula for this lithium aluminium magnesium dihydride and structures? Order of band gaps? BN > C(diamond) > AlP > InSb. The “bottom-up” approach requires one to “build up” nanoscale features from smaller entities. CHAPTER 25 Self-tests S25. 24. The Li and Al will be incorporated as metal hydride solid solutions. Li). As the temperature approaches 0 K.1 Synthesis of core-shell nanoparticles? The thermodynamic driving force is adjusted to a level that allows for heterogeneous nucleation of the shell material on the core but prevents homogeneous nucleation of the shell material. glass forming. 24.42 nm.2 Electron length and quantum confinement? A characteristic length is the exciton Bohr radius. it is a good candidate to consider. 24.3 Host for QDs? MCM-41. (BP)O4. the structure is very open. Zn. Ga. the sample needs to be made transparent to the electron beam. the shell could react with a specific location and the core could be used as a treatment. (b) Why are they used over other materials? They exhibit properties that are not observed in molecular or traditional solid state materials.17 Common features of self-assembly? (i) molecular or nanoscale subunits.12 PVD versus CVD? In CVD. an electron beam is transmitted through the materials. Results are much improved hardness values. eg. 25. such as nanorods of iron oxide. 25. (c) How are they made? Molecular beam epitaxy. (c) Stabilizers? Prevent surface oxidation and aggregation. large number of interfaces spaced on the nanoscale. (iii) growth of particles to desired size occurs. a liquid crystal. (iii) noncovalent interactions between the assembled subunits. (ii) spontaneous assembly of the subunits. (b) Vapor-phase.ANSWERS TO SELF-TESTS AND EXERCISES 47 allow for the precise spatial control over atoms and molecules. SEM sample preparation.. SEM versus TEM? SEM.9 Vapor-phase versus solution-based techniques? (a) Vapor-phase. more agglomeration. an oscillating chemical reaction.15 Superlattices and improved properties? AlN and TiN . TEM. (b) Why should the last two steps occur independently? So that nucleation fixes the total number of particles and growth leads to controlled size and narrow size distribution. the atomic species of interest are typically atomic. In TEM. (c) Purpose? In biosensing. 25. They can also be used to form quantum cascade lasers. 25.different elastic constants. then grow the particle in another. (iv) longer-range structures arising from the assembly process. In drug delivery. (ii) stable nuclei of nanometer dimensions formed. Also.13 (a) Purpose of QD layers? Multiple layers of quantum dots can increase the intensity of any optical absorption or emission. 25.7 25. In PVD. an electron beam is scanned over a material.11 (a) Homogeneous versus heterogeneous? Homogeneous nucleation leads to solid formation throughout the vapour phase. nanoparticles? Homogeneous nucleation is generally preferred for nanoparticle synthesis. their thermal energies are typically rather low.16 (a) Self-assembly? (a) Offers methods to bridge bottom-up and top-down approaches to synthesis. ensure the material is conductive.14 (a) Applications of quatum wells? Lasers and optical sensors. 25.8 (a) What are SPMs? A method to image the microscale features by scanning a very small probe over the surface and measuring some physical interaction between the tip and the material. the atomic species of interest are bound chemically to other species. 25. and an image is generated by recording the intensity of secondary or back-scattered electrons. 25.10 (a) What is a core-shell nanoparticle? 25. can be imaged using magnetic force microscopy. 25. (a) Steps in solutions synthesis of nanoparticles? (i) solvation.6 25. Dynamic self-assembly is when the system is oscillating between states and is dissipating energy in the process. and they limit traps for the holes and electrons. large sizes. (b) SPM and a specific nanomaterial? Local magnetic domains of magnetic nanomaterials. . Shell Core (b) How are they made? Nucleation in one solution. (b) Limitations? The limitations come from the requirements on how coherent the interface between the two materials must be. (b) In nanotechnology? Offers a route to assemble nanosized particles into macroscopic structures. eg. (b) thin film? Heterogeneous nucleation is (c) preferred in thin-film growth.18 Static versus dynamic self-assembly? Static self-assembly is when a system selfassembles to a stable state. 25. and the image is the spatial variation in the number of transmitted electrons. the dielectric property of the shell can control the surface plasmon of the core. and improve quantum yields and luminescence. (d) The hydrogenation of double bonds in vegetable oil? The process can be readily set up with existing technology. and exposed to pyridine vapor? complete dehydroxylation. A bionanomaterial and its application? PPF/PPF-DA is an injectable bionanomaterial used for bone-tissue engineering. Added phosphine will result in a lower concentration of the catalytically active 16electron complex. S26. (b) The hydrogenation of oil using a finely divided Ni catalyst? Heterogeneous.4 Self-tests S26.1 (c) The production of Li3N and its reaction with H2O? Not catalysis. cooled. block copolymers.19 25. Exercises 26.L Which of the following processes would be worth investigating? (a) The splitting of H2O into H2 and O2? Not worthwhile.4 Without R groups attached to the Zr center there is no preference for specific binding during polymerization. What is morphosynthesis? Control of nanoarchitectures in inorganic materials through changes in synthesis parameters. hybrid materials where no covalent or ionic bonds are present. (b) Selectivity? How much of the desired product is formed relative to by-products.. 26. Hybrid alumoxane nanoparticles dispersed in PPF/PPF-DA shows enhanced strength. 25. at least some of the components are linked through chemical bonds. Demonstate that the polymerization of propene with a simple Cp2ZrCl2 catalyst would give rise to atactic polypropene? D. CHAPTER 26 26.ANSWERS TO SELF-TESTS AND EXERCISES 48 25.20 25. polymer/clay nanocomposites. (a) Biomimetics? Designing nanomaterials that mimic biological systems.24. (a) Why is dispersion important in nanocomposites? They lead to increased exposed surface areas. S26. (c) Catalyst? A substance that increases the rate of a reaction but is not itself consumed.24)? No. (b) H2 plus O2 plus an electrical arc? Not catalysis.2 Define the following terms? (a) Turnover frequency? The amount of product formed per unit time per unit amount of catalyst. 26.5 Why does the addition of PPh3 to RhCl(PPh3)3 reduce the hydrogenation . (c) The conversion of D-glucose to a mixture by HCl? Homogeneous.2 γ-Alumina heated to 900ºC. (b) Why is dispersion difficult? The often nonpolar organic polymers do not have strong interactions with the polar or ionic inorganic components. (c) The combination of N2 with H2 to produce NH3? Very worthwhile reaction to try to catalyze efficiently at 80ºC. (b) Examples? Class I.3 Would a pure silica analog of ZSM-5 be an active catalyst for benzene alkylation (see Figure 26. S26. (b) example of biomimetics? Cellulose fibers in paper have been used to template the growth of titanium oxide nanotubes.1 Which of the following constitute catalysis? (a) H2 and C2H4 in contact with Pt? An example of genuine catalysis. Bionanocomposites and improved mechanical strength? Biomimetics need trabecular and cortical bone tissues.3 Classify the following as homogeneous or heterogeneous catalysis? (a) The increased rate of SO2 oxidation in the presence of NO? Homogeneous. (e) Catalyst support? Generally a ceramic like γ-alumina or silica gel. Class II. (d) Catalytic cycle? Sequence of chemical reactions involving the catalyst that transform the reactants into products. (a) What are the two classes of inorganicorganic nanocomposites? Class I.22 25. The effect of added phosphine on the catalytic activity of RhH(CO)(PPh3)3? (b) The decomposition of CO2 into C and O2? A waste of time.21 25. 26. Class II. which combine compressibility and tensile strength.23 25.25 Compare SAMs and cell membranes? SAM can structurally resemble a phospholipid bilayer. 7 26.6 What is the nature of binding at Cu blue centers as indicated by the EPR spectrum? There is greater covalence in blue Cu centers than in simple Cu(II) compounds. Co3+.11 Explain the trend in rates of H2 absorption by various olefins catalyzed by RhCl(PPh3)3? In both cases. .3-dimethylpentane? (i) The mechanism of deuterium exchange is probably related to the reverse of the last two reactions in Figure 26.10 Why might Cu sensors be designed to bind Cu(I) rather than Cu(II)? Cu(I) has an almost unique ability to undergo linear coordination by sulphur-containing ligands. the formation of either (A) or (E) is the rate-determining step. 26.ANSWERS TO SELF-TESTS AND EXERCISES 49 turnover frequency? The catalytic species that enters the cycle is RhCl(PPh3)2(Sol) (Sol = a solvent molecule).5 Why are iron-porphyrin complexes unable to bind O2 reversibly? The Fe(II) gets oxidized to Fe(III). CHAPTER 27 Self-tests S27.12 26.1 S27. and Fe3+.20.9 Suggest a reason why? (a) Ring opening alkene metathesis polymerization (ROMP) proceeds? ROMP can result in reduced steric strain.8 Suggest experiments that could establish the structure of the MoFe cofactor? EPR. 26.7 What is the nature of an active site with Copper (III)? Diamagnetic with squareplanar geometry. Unusual coordination of Mg? The protein’s 3D structure can place any particular atom in a suitable position for axial coordination. and EXAFS.13 Is Iron (II) expected to be present in the cell as uncomplexed ions? No.14 Why does CO decrease the effectiveness of Pt in catalyzing the reaction 2H+(aq) + 2e– → H2(g)? Platinum has a strong tendency to chemisorb CO. S27. S27.6 26. (a) Enhanced acidity? When Al3+ replaces Si4+ on lattice site charge is balanced by H3O+. Devise a plausible mechanism to explain the deuteration of 3. S27. (b) Three other ions? Ga3+.9 26. (b) Ring-closing metathesis (RCM) reaction proceeds? RCM results in the loss of ethane. (a) Attack by dissolved hydroxide? Structure C in Figure 26. single-crystal X-ray diffraction.10 26. S27. In the presence of added P(n-Bu)3. (ii) The second observation can be explained by invoking a mechanism for rapid deuterium S27.11 (b) Attack by coordinated hydroxide? Structure E given in Figure 26. 26.8 26. yielding an oxo-Iron(III) porphyrin complex. which can methylate anything in the cell.11 (c) Can one differentiate the stereochemistry? Yes. How does starting with MeCOOMe instead of MeOH lead to ethanoic anhydride instead of ethanoic acid using the Monsanto acetic acid process? The reaction of the ethanoate ion with the acetyl iodide leads to ethanoic anhydride. Hydroformylation catalysis with and without added P(n-Bu)3? The transformation of (E) into CoH(CO)4 must be the rate-determining step in the absence of added P(n-Bu)3. Why mercury is so toxic because of the action of enzymes containing cobalamin? Cobalamins are very active methyl transfer reactions. but is expensive. exchange of the methyl group in the chemisorbed –CHR(CH3) group (R = C(CH3)2(C2H5)). S27. 26. Why is the platinum-rhodium in automobile catalytic converters dispersed on the surface of a ceramic rather than used in the form of thin foil? A thin foil of platinum-rhodium will not have as much surface area as an equal amount of small particles finely dispersed on the surface of a ceramic support.3 Why does saline contain NaCl? Osmotic balance.4 Explain the significance of the Calcium ion pumps activation by calmodulin? The binding of calmodulin is a signal informing the pump that the cytoplasmic Ca2+ level has risen above a certain level. S27.15 Describe the role of an electrocatalyst? Platinum is the most efficient electrocatalyst for accelerating oxygen reduction at the fuel cell cathode. the alkene that is hydrogenated more slowly has a greater degree of substitution and so is sterically more demanding.2 S27. 2 Substituting Co2+ for Zn2+? Co(II) commonly adopts distorted tetrahedral and five-coordinate geometries typical of Zn(II) in enzymes.6 Explain the differences in the structures of the oxidized and reduced forms of the Pcluster in nitrogenase? The change in structure suggest that the coupling of proton and electron transfer can also occur at the Pcluster. and Mg(II)? strengths: Fe(III) > Zn(II) > Mg(II). 27. intermediate between the 3+ and the 2+ states.7 .3 Compare the acid/base catalytic activities Acid of Zn(II). by controlling protonation of the exchangeable ligands. 27. 27. enabling Zn enzymes to be studied by EPR.5 Interpret the Mossbauer spectra of ferredoxin? In the spectrum. Ligand binding rates are Mg(II) > Zn(II) > Fe(III). Fe(III). Fe(II) and Cu(I). What metals are involved in the synthesis of acetyl groups? Co(II). 27. with one pair spin-up and the other pair spin-down.1 Lanthanides versus calcium? They are hard Lewis acids and prefer coordination by hard bases.4 27. Gd3+ has excellent fluorescence. Co(II) is paramagnetic. 27.ANSWERS TO SELF-TESTS AND EXERCISES 50 Exercises 27. Propose a physical method for the EPR or determination of Fe(V)? Mossbauer. the oxidized spectrum is consistent with the iron atoms having the same valence.